diff options
| author | Anthony | 2020-05-24 20:59:14 -0500 |
|---|---|---|
| committer | Anthony | 2020-05-24 20:59:14 -0500 |
| commit | 7697dbf553b4b77a7953c86d3643a578b3d548a7 (patch) | |
| tree | bfcc8253a27c8f4a5edf34376f3da3345d0290e0 | |
| parent | ca235104c7f3516fb7f85d629d4910720842f492 (diff) | |
update
| -rw-r--r-- | .gitignore | 2 | ||||
| -rw-r--r-- | .replit | 2 | ||||
| -rw-r--r-- | cowntact.txt | 124 | ||||
| -rw-r--r-- | soc1.txt | 74 | ||||
| -rw-r--r-- | soc2.txt | 84 | ||||
| -rw-r--r-- | src/common.h | 8 | ||||
| -rw-r--r-- | src/huffman.cpp | 166 | ||||
| -rw-r--r-- | src/huffman.h | 10 | ||||
| -rw-r--r-- | src/main.cpp | 268 |
9 files changed, 369 insertions, 369 deletions
@@ -1,2 +1,2 @@ -out +out
.vscode
\ No newline at end of file @@ -1,2 +1,2 @@ -language = "cpp" +language = "cpp"
run = "g++ src/*.cpp -o out -std=c++11; ./out; rm out"
\ No newline at end of file diff --git a/cowntact.txt b/cowntact.txt index c834027..23fc638 100644 --- a/cowntact.txt +++ b/cowntact.txt @@ -1,62 +1,62 @@ -Farmer John is worried for the health of his cows (conveniently numbered -$1 ... N$ as always) after an outbreak of the highly contagious bovine -disease COWVID-19. - -Recently, Farmer John tested all of his cows and found some of them to be -positive for the disease. Using video footage from inside his barn, he is able -to review recent interactions between pairs of cows --- it turns out that when -cows greet each-other, they shake hooves, a gesture that can unfortunately -spread the infection from one cow to another. Farmer John assembles a -time-stamped list of interacting pairs of cows, with entries of the form -$(t, x, y)$, meaning that at time $t$, cow $x$ shook hooves with cow $y$. -Farmer John also knows the following: - -(i) Exactly one cow on his farm could have started out carrying the disease -(we'll call this cow "patient zero"). - -(ii) Once a cow is infected, she passes the infection along with her next $K$ -hoof shakes (possibly including the same partner cow several times). After -shaking hooves $K$ times, she no longer passes the infection along with -subsequent hoof shakes (since at this point she realizes she is spreading the -infection and washes her hooves carefully). - -(iii) Once a cow is infected, she stays infected. - -Unfortunately, Farmer John doesn't know which of his $N$ cows is patient zero, -nor does he know the value of $K$! Please help him narrow down the -possibilities for these unknowns based on his data. It is guaranteed that at -least one possibility is valid. - -INPUT FORMAT -The first line of the input file contains $N$ ($2 <= N <= 100$) and $T$ -($1 <= T <= 250$). The next line contains a string of length $N$ whose -entries are 0s and 1s, describing the current state of Farmer John's $N$ cows ---- 0 represents a healthy cow and 1 represents a cow presently with the -disease. Each of the next $T$ lines describes a record in Farmer John's list of -interactions and consists of three integers $t$, $x$, and $y$, where $t$ is a -positive integer time of the interaction ($t <= 250$) and $x$ and $y$ are -distinct integers in the range $1 \ldots N$, indicating which cows shook hands -at time $t$. At most one interaction happens at each point in time. - -OUTPUT FORMAT -Print a single line with three integers $x$, $y$, and $z$, where $x$ is the -number of possible cows who could have been patient zero, $y$ is the smallest -possible value of $K$ consistent with the data, and $z$ is the largest possible -value of $K$ consistent with the data (if there is no upper bound on $K$ that -can be deduced from the data, print "Infinity" for $z$). Note that it might be -possible to have $K=0$. - -SAMPLE INPUT -4 3 -1100 -7 1 2 -5 2 3 -6 2 4 - -SAMPLE OUTPUT -1 1 Infinity - -The only candidate for patient zero is cow 1. For all $K>0$, cow 1 infects cow 2 -at time 7, while cows 3 and 4 remain uninfected. - - +Farmer John is worried for the health of his cows (conveniently numbered
+$1 ... N$ as always) after an outbreak of the highly contagious bovine
+disease COWVID-19.
+
+Recently, Farmer John tested all of his cows and found some of them to be
+positive for the disease. Using video footage from inside his barn, he is able
+to review recent interactions between pairs of cows --- it turns out that when
+cows greet each-other, they shake hooves, a gesture that can unfortunately
+spread the infection from one cow to another. Farmer John assembles a
+time-stamped list of interacting pairs of cows, with entries of the form
+$(t, x, y)$, meaning that at time $t$, cow $x$ shook hooves with cow $y$.
+Farmer John also knows the following:
+
+(i) Exactly one cow on his farm could have started out carrying the disease
+(we'll call this cow "patient zero").
+
+(ii) Once a cow is infected, she passes the infection along with her next $K$
+hoof shakes (possibly including the same partner cow several times). After
+shaking hooves $K$ times, she no longer passes the infection along with
+subsequent hoof shakes (since at this point she realizes she is spreading the
+infection and washes her hooves carefully).
+
+(iii) Once a cow is infected, she stays infected.
+
+Unfortunately, Farmer John doesn't know which of his $N$ cows is patient zero,
+nor does he know the value of $K$! Please help him narrow down the
+possibilities for these unknowns based on his data. It is guaranteed that at
+least one possibility is valid.
+
+INPUT FORMAT
+The first line of the input file contains $N$ ($2 <= N <= 100$) and $T$
+($1 <= T <= 250$). The next line contains a string of length $N$ whose
+entries are 0s and 1s, describing the current state of Farmer John's $N$ cows
+--- 0 represents a healthy cow and 1 represents a cow presently with the
+disease. Each of the next $T$ lines describes a record in Farmer John's list of
+interactions and consists of three integers $t$, $x$, and $y$, where $t$ is a
+positive integer time of the interaction ($t <= 250$) and $x$ and $y$ are
+distinct integers in the range $1 \ldots N$, indicating which cows shook hands
+at time $t$. At most one interaction happens at each point in time.
+
+OUTPUT FORMAT
+Print a single line with three integers $x$, $y$, and $z$, where $x$ is the
+number of possible cows who could have been patient zero, $y$ is the smallest
+possible value of $K$ consistent with the data, and $z$ is the largest possible
+value of $K$ consistent with the data (if there is no upper bound on $K$ that
+can be deduced from the data, print "Infinity" for $z$). Note that it might be
+possible to have $K=0$.
+
+SAMPLE INPUT
+4 3
+1100
+7 1 2
+5 2 3
+6 2 4
+
+SAMPLE OUTPUT
+1 1 Infinity
+
+The only candidate for patient zero is cow 1. For all $K>0$, cow 1 infects cow 2
+at time 7, while cows 3 and 4 remain uninfected.
+
+
@@ -1,38 +1,38 @@ -A terrible new disease, COWVID-19, has begun to spread among cows worldwide. -Farmer John is trying to take as many precautions as possible to protect his -herd from infection. - -Farmer John's barn is a long narrow building containing $N$ stalls in a row -($2 <= N <= 10^5$). Some of these stalls are currently occupied by cows, -and some are vacant. Having read about the importance of "social distancing", -Farmer John wants to maximize $D$, where $D$ is the distance between the closest -two occupied stalls. For example, if stalls 3 and 8 are the closest that are -occupied, then $D = 5$. - -Two new cows recently joined Farmer John's herd and he needs to decide to which -formerly-unoccupied stalls they should be assigned. Please determine how he can -place his two new cows so that the resulting value of $D$ is still as large as -possible. Farmer John cannot move any of his existing cows; he only wants to -assign stalls to the new cows. - -INPUT FORMAT -The first line of input contains $N$. The next line contains a string of length -$N$ of 0s and 1s describing the sequence of stalls in the barn. 0s indicate -empty stalls and 1s indicate occupied stalls. The string has at least two 0s, -so there is at least enough room for two new cows. - -OUTPUT FORMAT -Please print the largest value of $D$ (the closest distance between two occupied -stalls) that Farmer John can achieve after adding his two new cows in an optimal -fashion. - -SAMPLE INPUT -14 -10001001000010 - -SAMPLE OUTPUT -2 - -In this example, Farmer John could add cows to make the occupancy string look -like 10x010010x0010, where x's indicate the new cows. In this case $D = 2$. It +A terrible new disease, COWVID-19, has begun to spread among cows worldwide.
+Farmer John is trying to take as many precautions as possible to protect his
+herd from infection.
+
+Farmer John's barn is a long narrow building containing $N$ stalls in a row
+($2 <= N <= 10^5$). Some of these stalls are currently occupied by cows,
+and some are vacant. Having read about the importance of "social distancing",
+Farmer John wants to maximize $D$, where $D$ is the distance between the closest
+two occupied stalls. For example, if stalls 3 and 8 are the closest that are
+occupied, then $D = 5$.
+
+Two new cows recently joined Farmer John's herd and he needs to decide to which
+formerly-unoccupied stalls they should be assigned. Please determine how he can
+place his two new cows so that the resulting value of $D$ is still as large as
+possible. Farmer John cannot move any of his existing cows; he only wants to
+assign stalls to the new cows.
+
+INPUT FORMAT
+The first line of input contains $N$. The next line contains a string of length
+$N$ of 0s and 1s describing the sequence of stalls in the barn. 0s indicate
+empty stalls and 1s indicate occupied stalls. The string has at least two 0s,
+so there is at least enough room for two new cows.
+
+OUTPUT FORMAT
+Please print the largest value of $D$ (the closest distance between two occupied
+stalls) that Farmer John can achieve after adding his two new cows in an optimal
+fashion.
+
+SAMPLE INPUT
+14
+10001001000010
+
+SAMPLE OUTPUT
+2
+
+In this example, Farmer John could add cows to make the occupancy string look
+like 10x010010x0010, where x's indicate the new cows. In this case $D = 2$. It
is impossible to add the new cows to achieve any higher value of $D$.
\ No newline at end of file @@ -1,43 +1,43 @@ -Farmer John is worried for the health of his cows after an outbreak of the -highly contagious bovine disease COWVID-19. - -Despite his best attempt at making his $N$ cows ($1 <= N <= 1000$) practice -"social distancing", many of them still unfortunately contracted the disease. -The cows, conveniently numbered $1 \ldots N$, are each standing at distinct -points along a long path (essentially a one-dimensional number line), with cow -$i$ standing at position $x_i$. Farmer John knows that there is a radius $R$ -such that any cow standing up to and including $R$ units away from an infected -cow will also become infected (and will then pass the infection along to -additional cows within $R$ units away, and so on). - -Unfortunately, Farmer John doesn't know $R$ exactly. He does however know which -of his cows are infected. Given this data, please determine the minimum -possible number of cows that were initially infected with the disease. - -INPUT FORMAT -The first line of input contains $N$. The next $N$ lines each describe one cow -in terms of two integers, $x$ and $s$, where $x$ is the position -($0 <= x <= 10^6$), and $s$ is 0 for a healthy cow or 1 for a sick cow. At -least one cow is sick, and all cows that could possibly have become sick from -spread of the disease have now become sick. - -OUTPUT FORMAT -Please output the minimum number of cows that could have initially been sick, -prior to any spread of the disease. - -SAMPLE INPUT -6 -7 1 -1 1 -15 1 -3 1 -10 0 -6 1 - -SAMPLE OUTPUT -3 - -In this example, we know that $R < 3$ since otherwise the cow at position 7 -would have infected the cow at position 10. Therefore, at least 3 cows must -have started out infected -- one of the two cows at positions 1 and 3, one of +Farmer John is worried for the health of his cows after an outbreak of the
+highly contagious bovine disease COWVID-19.
+
+Despite his best attempt at making his $N$ cows ($1 <= N <= 1000$) practice
+"social distancing", many of them still unfortunately contracted the disease.
+The cows, conveniently numbered $1 \ldots N$, are each standing at distinct
+points along a long path (essentially a one-dimensional number line), with cow
+$i$ standing at position $x_i$. Farmer John knows that there is a radius $R$
+such that any cow standing up to and including $R$ units away from an infected
+cow will also become infected (and will then pass the infection along to
+additional cows within $R$ units away, and so on).
+
+Unfortunately, Farmer John doesn't know $R$ exactly. He does however know which
+of his cows are infected. Given this data, please determine the minimum
+possible number of cows that were initially infected with the disease.
+
+INPUT FORMAT
+The first line of input contains $N$. The next $N$ lines each describe one cow
+in terms of two integers, $x$ and $s$, where $x$ is the position
+($0 <= x <= 10^6$), and $s$ is 0 for a healthy cow or 1 for a sick cow. At
+least one cow is sick, and all cows that could possibly have become sick from
+spread of the disease have now become sick.
+
+OUTPUT FORMAT
+Please output the minimum number of cows that could have initially been sick,
+prior to any spread of the disease.
+
+SAMPLE INPUT
+6
+7 1
+1 1
+15 1
+3 1
+10 0
+6 1
+
+SAMPLE OUTPUT
+3
+
+In this example, we know that $R < 3$ since otherwise the cow at position 7
+would have infected the cow at position 10. Therefore, at least 3 cows must
+have started out infected -- one of the two cows at positions 1 and 3, one of
the two cows at positions 6 and 7, and the cow at position 15.
\ No newline at end of file diff --git a/src/common.h b/src/common.h index c295a6a..7859893 100644 --- a/src/common.h +++ b/src/common.h @@ -1,5 +1,5 @@ -// include this file at the top of every source file - -#pragma once -#include <bits/stdc++.h> +// include this file at the top of every source file
+
+#pragma once
+#include <bits/stdc++.h>
using namespace std;
\ No newline at end of file diff --git a/src/huffman.cpp b/src/huffman.cpp index dfe36cb..1644284 100644 --- a/src/huffman.cpp +++ b/src/huffman.cpp @@ -1,84 +1,84 @@ -#include "huffman.h" -#include "common.h" - -namespace huffman { - struct node { - char c; int f; node * l, * r; - node(char _c, int _f) { c = _c, f = _f, l = r = NULL; } - }; - struct comp { bool operator()(node * a, node * b) { return a->f > b->f; } }; - node * root; - vector<bool> path, code[128]; - void traverse(node * n) { - if (n->l) path.push_back(0), traverse(n->l), path.pop_back(); - if (n->r) path.push_back(1), traverse(n->r), path.pop_back(); - if (n->c) code[n->c] = path; - } - - void encode_tree(node * n, vector<bool> & v) { - if (n->c) { - v.push_back(1); - for (int i = 0; i < 8; ++i) v.push_back(1 & (n->c >> i)); - } - else { - v.push_back(0); - if (n->l) v.push_back(1), encode_tree(n->l, v); - else v.push_back(0); - if (n->r) v.push_back(1), encode_tree(n->r, v); - else v.push_back(0); - } - } - - int idx = 0; - void decode_tree(node * n, vector<bool> & v) { - if (v[idx++] == 1) { - for (int i = 0; i < 8; ++i) if (v[idx++]) n->c |= (1 << i); - } - else { - if (v[idx++] == 1) n->l = new node(0, 0), decode_tree(n->l, v); - if (v[idx++] == 1) n->r = new node(0, 0), decode_tree(n->r, v); - } - } - - void generate(vector<int> f) { - priority_queue<node *, vector<node *>, comp> pq; - for (int c = 1; c < 128; ++c) { - node * n = new node(c, f[c]); - pq.push(n); - } - while (pq.size() > 1) { - node * l = pq.top(); pq.pop(); node * r = pq.top(); pq.pop(); - node * n = new node(0, l->f + r->f); - n->l = l, n->r = r; - pq.push(n); - } - root = pq.top(); - } - - void solve(node * n, vector<bool> & v, string & s) { - if (idx > v.size()) return; - if (n->c) s += n->c, solve(root, v, s); - else v[idx++] == 0 ? solve(n->l, v, s) : solve(n->r, v, s); - } - - vector<bool> encode(string s) { - vector<int> f(128, 0); - for (auto& c : s) ++f[c]; - generate(f); - traverse(root); - vector<bool> ret; - encode_tree(root, ret); - for (auto& c : s) { - for (auto b : code[c]) ret.push_back(b); - } - return ret; - } - - string decode(vector<bool> v) { - root = new node(0, 0); - decode_tree(root, v); - string ret; - solve(root, v, ret); - return ret; - } +#include "huffman.h"
+#include "common.h"
+
+namespace huffman {
+ struct node {
+ char c; int f; node * l, * r;
+ node(char _c, int _f) { c = _c, f = _f, l = r = NULL; }
+ };
+ struct comp { bool operator()(node * a, node * b) { return a->f > b->f; } };
+ node * root;
+ vector<bool> path, code[128];
+ void traverse(node * n) {
+ if (n->l) path.push_back(0), traverse(n->l), path.pop_back();
+ if (n->r) path.push_back(1), traverse(n->r), path.pop_back();
+ if (n->c) code[n->c] = path;
+ }
+
+ void encode_tree(node * n, vector<bool> & v) {
+ if (n->c) {
+ v.push_back(1);
+ for (int i = 0; i < 8; ++i) v.push_back(1 & (n->c >> i));
+ }
+ else {
+ v.push_back(0);
+ if (n->l) v.push_back(1), encode_tree(n->l, v);
+ else v.push_back(0);
+ if (n->r) v.push_back(1), encode_tree(n->r, v);
+ else v.push_back(0);
+ }
+ }
+
+ int idx = 0;
+ void decode_tree(node * n, vector<bool> & v) {
+ if (v[idx++] == 1) {
+ for (int i = 0; i < 8; ++i) if (v[idx++]) n->c |= (1 << i);
+ }
+ else {
+ if (v[idx++] == 1) n->l = new node(0, 0), decode_tree(n->l, v);
+ if (v[idx++] == 1) n->r = new node(0, 0), decode_tree(n->r, v);
+ }
+ }
+
+ void generate(vector<int> f) {
+ priority_queue<node *, vector<node *>, comp> pq;
+ for (int c = 1; c < 128; ++c) {
+ node * n = new node(c, f[c]);
+ pq.push(n);
+ }
+ while (pq.size() > 1) {
+ node * l = pq.top(); pq.pop(); node * r = pq.top(); pq.pop();
+ node * n = new node(0, l->f + r->f);
+ n->l = l, n->r = r;
+ pq.push(n);
+ }
+ root = pq.top();
+ }
+
+ void solve(node * n, vector<bool> & v, string & s) {
+ if (idx > v.size()) return;
+ if (n->c) s += n->c, solve(root, v, s);
+ else v[idx++] == 0 ? solve(n->l, v, s) : solve(n->r, v, s);
+ }
+
+ vector<bool> encode(string s) {
+ vector<int> f(128, 0);
+ for (auto& c : s) ++f[c];
+ generate(f);
+ traverse(root);
+ vector<bool> ret;
+ encode_tree(root, ret);
+ for (auto& c : s) {
+ for (auto b : code[c]) ret.push_back(b);
+ }
+ return ret;
+ }
+
+ string decode(vector<bool> v) {
+ root = new node(0, 0);
+ decode_tree(root, v);
+ string ret;
+ solve(root, v, ret);
+ return ret;
+ }
}
\ No newline at end of file diff --git a/src/huffman.h b/src/huffman.h index 707b74d..4754ca0 100644 --- a/src/huffman.h +++ b/src/huffman.h @@ -1,6 +1,6 @@ -#include "common.h" - -namespace huffman { - vector<bool> encode(string s); - string decode(vector<bool> v); +#include "common.h"
+
+namespace huffman {
+ vector<bool> encode(string s);
+ string decode(vector<bool> v);
}
\ No newline at end of file diff --git a/src/main.cpp b/src/main.cpp index 904375d..ce7c28b 100644 --- a/src/main.cpp +++ b/src/main.cpp @@ -1,135 +1,135 @@ -#include "common.h" -#include "huffman.h" -#define pb push_back - - - -int getweight(string s) -{ - if(s[0]=='$') - return 1000; - else if(s=="farmer"||s.length()==1||s=="example")return 1; - else return 10; -} -struct clause -{ - vector<string> words; - vector<int> weights; - string s; - int weight; - void parse() - { - string word=""; - bool iseq=false; - for(int j=0;j<s.length();j++) - { - if(s.at(j)=='$') iseq=!iseq; - if(('a'<=s.at(j)&&s.at(j)<='z')||('A'<=s.at(j)&&s.at(j)<='Z')||('0'<=s.at(j)&&s.at(j)<='9')||(iseq&&s.at(j)!=' ')) - word+=('A'<=s.at(j)&&s.at(j)<='Z')?(s.at(j)-'A'+'a'):(s.at(j)); - else - { - if(!iseq) - { - if(word!="")words.pb(word); - word=""; - } - } - } - if(word!="") words.pb(word); - } - void calcweight() - { - for(auto& w: words)weights.pb(getweight(w)); - weight=-100000; - for(auto& w : weights)if(w>weight)weight=w; - } -}; -struct sentence -{ - vector<clause> clauses; - string s; - int weight; - void parse() - { - string cl=""; - bool iseq=false; - for(int j=0;j<s.length();j++) - { - if(s.at(j)=='$') iseq=!iseq; - if(s.at(j)!=','|| iseq) cl+=s.at(j); - else - { - clause tmp; - tmp.s=cl;clauses.pb(tmp);cl=""; - } - } - clause tmp; - tmp.s=cl;clauses.pb(tmp); - for(auto& c:clauses) c.parse(); - int numclause=0; - for(int i=0;i<clauses.size();i++) - if(clauses[i].words.size()!=0) - { - auto tmp=clauses[i]; - clauses[numclause]=tmp; - numclause++; - } - clauses.resize(numclause); - } - void calcweight() - { - weight=-100000; - for(auto& c : clauses) - { - c.calcweight(); - if(c.weight>weight)weight=c.weight; - } - } -}; -vector<sentence> text; -vector<int> value; -string line; -int b,c; -void init() -{ - cin>>c; - ifstream stin("soc1.txt"); - string sentencefile=""; - bool iseq=false; - while(getline(stin,line)) - { - for(int i=0;i<line.length();i++) - { - b++; - if(line.at(i)=='$') iseq=!iseq; - if((line.at(i)!='.'&&line.at(i)!='?'&&line.at(i)!='!')|| iseq)sentencefile+=line.at(i); - else - { - sentence tmp;tmp.s=sentencefile;text.pb(tmp); - sentencefile=""; - } - } - b++; - sentencefile+=' '; - } - for(auto& i:text)i.parse(); -} -int main() { - /*init(); - for(sentence& i:text) - { - for(clause& j:i.clauses) - { - for(auto k:j.words) - { - cout<<k<<endl; - } - } - }*/ - - vector<bool> test = huffman::encode("asdfjkl;1234"); - for (auto b : test) cout << b; - cout << '\n'; - string s = huffman::decode(test); - cout << s << '\n'; +#include "common.h"
+#include "huffman.h"
+#define pb push_back
+
+
+
+int getweight(string s)
+{
+ if(s[0]=='$')
+ return 1000;
+ else if(s=="farmer"||s.length()==1||s=="example")return 1;
+ else return 10;
+}
+struct clause
+{
+ vector<string> words;
+ vector<int> weights;
+ string s;
+ int weight;
+ void parse()
+ {
+ string word="";
+ bool iseq=false;
+ for(int j=0;j<s.length();j++)
+ {
+ if(s.at(j)=='$') iseq=!iseq;
+ if(('a'<=s.at(j)&&s.at(j)<='z')||('A'<=s.at(j)&&s.at(j)<='Z')||('0'<=s.at(j)&&s.at(j)<='9')||(iseq&&s.at(j)!=' '))
+ word+=('A'<=s.at(j)&&s.at(j)<='Z')?(s.at(j)-'A'+'a'):(s.at(j));
+ else
+ {
+ if(!iseq)
+ {
+ if(word!="")words.pb(word);
+ word="";
+ }
+ }
+ }
+ if(word!="") words.pb(word);
+ }
+ void calcweight()
+ {
+ for(auto& w: words)weights.pb(getweight(w));
+ weight=-100000;
+ for(auto& w : weights)if(w>weight)weight=w;
+ }
+};
+struct sentence
+{
+ vector<clause> clauses;
+ string s;
+ int weight;
+ void parse()
+ {
+ string cl="";
+ bool iseq=false;
+ for(int j=0;j<s.length();j++)
+ {
+ if(s.at(j)=='$') iseq=!iseq;
+ if(s.at(j)!=','|| iseq) cl+=s.at(j);
+ else
+ {
+ clause tmp;
+ tmp.s=cl;clauses.pb(tmp);cl="";
+ }
+ }
+ clause tmp;
+ tmp.s=cl;clauses.pb(tmp);
+ for(auto& c:clauses) c.parse();
+ int numclause=0;
+ for(int i=0;i<clauses.size();i++)
+ if(clauses[i].words.size()!=0)
+ {
+ auto tmp=clauses[i];
+ clauses[numclause]=tmp;
+ numclause++;
+ }
+ clauses.resize(numclause);
+ }
+ void calcweight()
+ {
+ weight=-100000;
+ for(auto& c : clauses)
+ {
+ c.calcweight();
+ if(c.weight>weight)weight=c.weight;
+ }
+ }
+};
+vector<sentence> text;
+vector<int> value;
+string line;
+int b,c;
+void init()
+{
+ cin>>c;
+ ifstream stin("soc1.txt");
+ string sentencefile="";
+ bool iseq=false;
+ while(getline(stin,line))
+ {
+ for(int i=0;i<line.length();i++)
+ {
+ b++;
+ if(line.at(i)=='$') iseq=!iseq;
+ if((line.at(i)!='.'&&line.at(i)!='?'&&line.at(i)!='!')|| iseq)sentencefile+=line.at(i);
+ else
+ {
+ sentence tmp;tmp.s=sentencefile;text.pb(tmp);
+ sentencefile="";
+ }
+ }
+ b++;
+ sentencefile+=' ';
+ }
+ for(auto& i:text)i.parse();
+}
+int main() {
+ /*init();
+ for(sentence& i:text)
+ {
+ for(clause& j:i.clauses)
+ {
+ for(auto k:j.words)
+ {
+ cout<<k<<endl;
+ }
+ }
+ }*/
+
+ vector<bool> test = huffman::encode("asdfjkl;1234");
+ for (auto b : test) cout << b;
+ cout << '\n';
+ string s = huffman::decode(test);
+ cout << s << '\n';
}
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