题意：

有2*n把钥匙配成n对，每对中只能使用一把，另外有m道门，每道门能被2把药匙打开，问最多能从1开始按顺序打开多少道门。

分析：

二分枚举能打开的门数，用2-sat算法判断能否打开。

代码：

//poj 2723
//sep9
#include <iostream>
#include <cstdio>
#include <string.h>
#include <stack>
using namespace std;
const int maxN=10024;
const int maxM=100024;
int e,n,m,t,ecnt;
int head[maxN],ins[maxN],low[maxN],dfn[maxN];
int sol[maxN],belong[maxN];
stack<int> s;
struct Edge
{
	int v,next;
}edge[maxM];
struct Door
{
	int x,y;
}door[maxM]; 
struct Lock
{
	int x,y;
}lock[maxN];
void addegde(int u,int v)
{
	edge[e].v=v;edge[e].next=head[u];
	head[u]=e++;		
}

void dfs(int x)
{
	low[x]=dfn[x]=++t;	
	s.push(x);
	ins[x]=1;
	for(int i=head[x];i!=-1;i=edge[i].next){
		int v=edge[i].v;
		if(!dfn[v]){
			dfs(v);
			low[x]=min(low[x],low[v]);
		}else if(ins[v]==1)
			low[x]=min(low[x],dfn[v]);
	} 
	if(dfn[x]==low[x]){
		++ecnt;		
		int k;
		do{
			k=s.top();
			s.pop();
			ins[k]=0;
			belong[k]=ecnt;
		}while(dfn[k]!=low[k]);
	}
}
int two_sat()
{
	memset(ins,0,sizeof(ins));	
	memset(dfn,0,sizeof(dfn)); 
	while(!s.empty()) s.pop();
	int i;
	t=0,ecnt=0;
	for(i=1;i<=2*n;++i)
		if(!dfn[i])
			dfs(i);
	for(i=1;i<=n;++i)
		if(belong[i]==belong[i+n])
			return 0;
	return 1;
}

int pass(int mid)
{
	if(mid==0)
		return 1;
	int i;
	e=0;
	memset(head,-1,sizeof(head));
	for(i=1;i<=n;++i){
		int a=lock[i].x;
		int b=lock[i].y;
		addegde(a,b+n);
		addegde(b,a+n);
	}
		
	for(i=1;i<=mid;++i){
		int a=door[i].x;
		int b=door[i].y;	
		addegde(a+n,b);
		addegde(b+n,a);
	}			
	return two_sat();
}

int main()
{
	while(scanf("%d%d",&n,&m)==2){
		if(m==0&&n==0)
			break; 
		int i;
		for(i=1;i<=n;++i){
			scanf("%d%d",&lock[i].x,&lock[i].y);
			++lock[i].x;
			++lock[i].y;
		}
		n*=2;
 		for(i=1;i<=m;++i){
 			scanf("%d%d",&door[i].x,&door[i].y);
 			++door[i].x;
 			++door[i].y;
		 }
		int l=0,r=m+1;
		while(l<r){
			int mid=(r+l)/2;
			if(pass(mid))
				l=mid+1;
			else
				r=mid;	
		} 
		printf("%d\n",l-1);
	}
	return 0;	
}

poj 2723 Get Luffy Out 2-sat