POJ 3678 Katu Puzzle

Katu Puzzle

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on  PKU. Original ID: 3678
64-bit integer IO format: %lld      Java class name: Main

 

Katu Puzzle is presented as a directed graph G(V, E) with each edge e(a, b) labeled by a boolean operator op (one of AND, OR, XOR) and an integer c (0 ≤ c ≤ 1). One Katu is solvable if one can find each vertex Vi a value Xi (0 ≤ Xi ≤ 1) such that for each edge e(a, b) labeled by op and c, the following formula holds:

 Xa op Xb = c

The calculating rules are:

AND 0 1 0 0 0 1 0 1

OR 0 1 0 0 1 1 1 1

XOR 0 1 0 0 1 1 1 0

Given a Katu Puzzle, your task is to determine whether it is solvable.

 

Input

The first line contains two integers N (1 ≤ N ≤ 1000) and M,(0 ≤ M ≤ 1,000,000) indicating the number of vertices and edges.The following M lines contain three integers a (0 ≤ a <N), b(0 ≤ b < N), c and an operator op each, describing the edges.

 

Output

Output a line containing "YES" or "NO".

 

Sample Input

4 4

0 1 1 AND

1 2 1 OR

3 2 0 AND

3 0 0 XOR

Sample Output

YES

Source

POJ Founder Monthly Contest – 2008.07.27

 

解题：2SAT

 

  1 #include <iostream>

  2 #include <cstdio>

  3 #include <cstring>

  4 #include <cmath>

  5 #include <algorithm>

  6 #include <climits>

  7 #include <vector>

  8 #include <queue>

  9 #include <cstdlib>

 10 #include <string>

 11 #include <set>

 12 #include <stack>

 13 #define LL long long

 14 #define pii pair<int,int>

 15 #define INF 0x3f3f3f3f

 16 using namespace std;

 17 const int maxn = 2010;

 18 struct arc{

 19     int to,next;

 20     arc(int x = 0,int y = -1){

 21         to = x;

 22         next = y;

 23     }

 24 };

 25 arc e[4000010];

 26 int head[maxn],dfn[maxn],low[maxn],belong[maxn];

 27 int tot,cnt,scc,n,m;

 28 bool instack[maxn];

 29 stack<int>stk;

 30 void add(int u,int v){

 31     e[tot] = arc(v,head[u]);

 32     head[u] = tot++;

 33 }

 34 void init(){

 35     for(int i = 0; i < maxn; i++){

 36         belong[i] = 0;

 37         low[i] = dfn[i] = 0;

 38         head[i] = -1;

 39         instack[i] = false;

 40     }

 41     while(!stk.empty()) stk.pop();

 42     tot = cnt = scc = 0;

 43 }

 44 void tarjan(int u){

 45     dfn[u] = low[u] = ++cnt;

 46     instack[u] = true;

 47     stk.push(u);

 48     for(int i = head[u]; ~i; i = e[i].next){

 49         if(!dfn[e[i].to]){

 50             tarjan(e[i].to);

 51             if(low[e[i].to] < low[u]) low[u] = low[e[i].to];

 52         }else if(instack[e[i].to] && dfn[e[i].to] < low[u])

 53             low[u] = dfn[e[i].to];

 54     }

 55     if(dfn[u] == low[u]){

 56         scc++;

 57         int v;

 58         do{

 59             v = stk.top();

 60             stk.pop();

 61             instack[v] = false;

 62             belong[v] = scc;

 63         }while(v != u);

 64     }

 65 }

 66 bool solve(){

 67     for(int i = 0; i < (n<<1); i++)

 68         if(!dfn[i]) tarjan(i);

 69     for(int i = 0; i < n; i++)

 70         if(belong[i] == belong[i+n]) return false;

 71     return true;

 72 }

 73 int main() {

 74     char op[5];

 75     int u,v,c;

 76     while(~scanf("%d %d",&n,&m)){

 77         init();

 78         while(m--){

 79             scanf("%d %d %d %s",&u,&v,&c,op);

 80             if(op[0] == 'A'){

 81                 if(c){

 82                     add(u+n,v+n);

 83                     add(v+n,u+n);

 84                     add(u,u+n);

 85                     add(v,v+n);

 86                 }else{

 87                     add(u+n,v);

 88                     add(v+n,u);

 89                 }

 90             }else if(op[0] == 'O'){

 91                 if(c){

 92                     add(u,v+n);

 93                     add(v,u+n);

 94                 }else{

 95                     add(u,v);

 96                     add(v,u);

 97                     add(u+n,u);

 98                     add(v+n,v);

 99                 }

100             }else if(op[0] == 'X'){

101                 if(c){

102                     add(u,v+n);

103                     add(v,u+n);

104                     add(u+n,v);

105                     add(v+n,u);

106                 }else{

107                     add(u,v);

108                     add(v,u);

109                     add(u+n,v+n);

110                     add(v+n,u+n);

111                 }

112             }

113         }

114         printf("%s\n",solve()?"YES":"NO");

115     }

116     return 0;

117 }

View Code