Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 6893 | Accepted: 2531 |
Description
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
Hint
Source
#include<cstdio> #include<cstring> #include<queue> #include<vector> #include<algorithm> #include<iostream> using namespace std; const int maxn = 1000 + 5; struct TwoSAT { int n; vector<int> G[maxn*2]; bool mark[maxn*2]; int S[maxn*2], c; bool dfs(int x) { if (mark[x^1]) return false; if (mark[x]) return true; mark[x] = true; S[c++] = x; for (int i = 0; i < G[x].size(); i++) if (!dfs(G[x][i])) return false; return true; } void init(int n) { this->n = n; for (int i = 0; i < n*2; i++) G[i].clear(); memset(mark, 0, sizeof(mark)); } //kind=0 -> &&;kind=1 -> ||;kind=2 -> ^; void add_edge(int a,int b,int c,int kind){ if(kind == 0){ if(c == 0){ G[2*a].push_back(2*b+1); G[2*b].push_back(2*a+1); } else{ G[2*a+1].push_back(2*a); G[2*b+1].push_back(2*b); } } else if(kind == 1){ if(c == 0){ G[2*a].push_back(2*a+1); G[2*b].push_back(2*b+1); } else{ G[2*a+1].push_back(2*b); G[2*b+1].push_back(2*a); } } else{ if(c == 0){ G[2*a].push_back(2*b); G[2*a+1].push_back(2*b+1); G[2*b].push_back(2*a); G[2*b+1].push_back(2*a+1); } else{ G[2*a].push_back(2*b+1); G[2*a+1].push_back(2*b); G[2*b].push_back(2*a+1); G[2*b+1].push_back(2*a); } } } bool solve() { for(int i = 0; i < n*2; i += 2){ if(!mark[i] && !mark[i+1]) { c = 0; if(!dfs(i)) { while(c > 0) mark[S[--c]] = false; if(!dfs(i+1)) return false; } } } return true; } }; TwoSAT solver; int main(){ int n,m; while(scanf("%d%d",&n,&m) != EOF){ solver.init(n); for(int i = 0;i < m;i++){ int x,y,val; char s[5]; scanf("%d%d%d%s",&x,&y,&val,s); if(s[0] == 'A') solver.add_edge(x,y,val,0); else if(s[0] == 'O') solver.add_edge(x,y,val,1); else solver.add_edge(x,y,val,2); } if(solver.solve()) printf("YES\n"); else printf("NO\n"); } return 0; }模板解释,每个点拆成2*i(真)和2*i+1(假),下标从0开始。加边函数参数,kind代表条件的类型,c是a op b 的值。