Katu Puzzle
Time Limit: 1000ms
Memory Limit: 65536KB
This problem will be judged on PKU. Original ID: 367864-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:
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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
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 ",solve()?"YES":"NO"); 115 } 116 return 0; 117 }