传送门：点击打开链接
题意：n对夫妻，一对夫妻必须去1人，有m对人互相讨厌，不能同时去，问是否有这样的情况。
思路：2SAT...
#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <bitset>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define fuck(x) cout<<"["<<x<<"]";
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w+",stdout);
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;const int MX = 2e3 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;struct Edge {int v, nxt;
} E[MX * MX * 2];
int Head[MX][2], erear;
void edge_init() {erear = 0;memset(Head, -1, sizeof(Head));
}
void edge_add(int z, int u, int v) {E[erear].v = v;E[erear].nxt = Head[u][z];Head[u][z] = erear++;
}
void edge_add(int u, int v) {edge_add(0, u, v);edge_add(1, v, u);
}
int Stack[MX], Belong[MX], vis[MX], ssz, bsz;
void DFS(int u, int s) {vis[u] = 1;if(s) Belong[u] = s;for(int i = Head[u][s > 0]; ~i; i = E[i].nxt) {int v = E[i].v;if(!vis[v]) DFS(v, s);}if(!s) Stack[++ssz] = u;
}
void tarjan(int n) {ssz = bsz = 0;for(int i = 1; i <= n; i++) vis[i] = 0;for(int i = 1; i <= n; i++) {if(!vis[i]) DFS(i, 0);}for(int i = 1; i <= n; i++) vis[i] = 0;for(int i = ssz; i >= 1; i--) {if(!vis[Stack[i]]) DFS(Stack[i], ++bsz);}
}int main() {int n, m; //FIN;while(~scanf("%d%d", &n, &m)) {edge_init();for(int i = 1; i <= m; i++) {int u, v, c1, c2;scanf("%d%d%d%d", &u, &v, &c1, &c2);u++; v++;edge_add(u + (c1 ? n : 0), v + (c2 ? 0 : n));edge_add(v + (c2 ? n : 0), u + (c1 ? 0 : n));}tarjan(2 * n);bool ans = true;for(int i = 1; i <= n; i++) {if(Belong[i] == Belong[i + n]) {ans = false; break;}}printf("%s\n", ans ? "YES" : "NO");}return 0;
}