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2-sat模板题

判断是否存在可行解，建图后求强连通分量逐一判断所有对立点是否都在不同强连通分量

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
#define N 2010
#define M 8000010  //2 * N * N
#define cl memset
#define pb push_back

int n,tot,head[N];
struct edge{
	int v,next;
}e[M];
int dfn[N],low[N],belong[N],stack[N],dcnt,bcnt,top;
bool ins[N];

inline void add(int u ,int v){
	e[tot].v = v; e[tot].next = head[u]; head[u] = tot++;
}

void tarjan(int u){
	dfn[u] = low[u] = ++dcnt;
	stack[++top] = u; ins[u] = true;
	for(int k = head[u]; k != -1; k = e[k].next){
		int v = e[k].v;
		if(!dfn[v]){
			tarjan(v);
			low[u] = min(low[u],low[v]);
		}
		else if(ins[v])
			low[u] = min(low[u],dfn[v]);
	}
	if(dfn[u] == low[u]){
		++bcnt;
		while(true){
			int x = stack[top--];
			ins[x] = false;
			belong[x] = bcnt;
			if(x == u) break;
		}
	}
}

void scc(){
	dcnt = bcnt = top = 0;
	cl(dfn,0,sizeof(dfn));
	cl(ins,false,sizeof(ins));
	for(int i = 0; i < 2*n; i++)
		if(!dfn[i])
			tarjan(i);
}

int main(){
	while(scanf("%d",&n)!=EOF){
		int m;
		scanf("%d",&m);
		tot = 0;
		cl(head,-1,sizeof(head));
		while(m--){
			int x,y,fx,fy;
			scanf("%d%d%d%d",&x,&y,&fx,&fy);
			add((x<<1) + fx , (y<<1|1) - fy);
			add((y<<1) + fy , (x<<1|1) - fx);
		}
		scc();
		bool res = true;
		for(int i = 0; i < n ; i++)
			if(belong[i<<1] == belong[i<<1|1]){
				res = false; break;
			}
		if(res) puts("YES");
		else	puts("NO");
	}
	return 0;
}