http://acm.hdu.edu.cn/showproblem.php?pid=3062
题意：中文。
思路：裸的2-SAT。判断二元组的两个人是否在同一个强连通分量。
学习地址：http://www.cnblogs.com/ambition/archive/2011/07/30/2-sat.html
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define N 2010 4 struct Edge { 5 int u, v, nxt; 6 } edge[N*N*2]; 7 int head[N], tot, belong[N], num, dfn[N], low[N], vis[N], tid; 8 stack<int> sta; 9 10 void Add(int u, int v) { 11 edge[tot] = (Edge) { u, v, head[u] }; head[u] = tot++; 12 } 13 14 void init() { 15 memset(head, -1, sizeof(head)); 16 memset(vis, 0, sizeof(vis)); 17 memset(dfn, 0, sizeof(dfn)); 18 memset(low, 0, sizeof(low)); 19 tot = num = tid = 0; 20 while(!sta.empty()) sta.pop(); 21 } 22 23 void tarjan(int u) { 24 dfn[u] = low[u] = ++tid; 25 vis[u] = 1; sta.push(u); 26 for(int i = head[u]; ~i; i = edge[i].nxt) { 27 int v = edge[i].v; 28 if(!dfn[v]) { 29 tarjan(v); 30 if(low[u] > low[v]) low[u] = low[v]; 31 } else if (vis[v]) { 32 if(low[u] > dfn[v]) low[u] = dfn[v]; 33 } 34 } 35 if(low[u] == dfn[u]) { 36 num++; 37 while(true) { 38 int x = sta.top(); sta.pop(); 39 belong[x] = num; 40 vis[x] = 0; 41 if(x == u) break; 42 } 43 } 44 } 45 46 int main() { 47 int n, m; 48 while(~scanf("%d", &n)) { 49 scanf("%d", &m); init(); 50 for(int i = 0; i < m; i++) { 51 int a, b, c, d; 52 scanf("%d%d%d%d", &a, &b, &c, &d); 53 Add(a * 2 + c, b * 2 + (d + 1) % 2); 54 Add(b * 2 + d, a * 2 + (c + 1) % 2); 55 } 56 for(int i = 0; i < 2 * n; i++) if(!dfn[i]) tarjan(i); 57 bool flag = 1; 58 for(int i = 0; i < n; i++) 59 if(belong[i*2] == belong[i*2+1]) flag = 0; 60 if(flag) puts("YES"); 61 else puts("NO"); 62 } 63 return 0; 64 }