POJ 2723 Get Luffy Out（2-SAT）

#include <cstdio>
#include <algorithm>
#include <vector>
#include <cstring> 
using namespace std;
const int MAX_V=10000;
int V; //顶点数
vector<int> G[MAX_V]; //图的邻接表表示
vector<int> rG[MAX_V]; //反向图
vector<int> vs; //后序遍历
bool used[MAX_V];
int cmp[MAX_V]; //所属强连通分量的拓扑序
void add_edge(int from,int to){
    G[from].push_back(to);
    rG[to].push_back(from);
} 
void dfs(int v){
    used[v]=1;
    for(int i=0;i<G[v].size();i++){
        if(!used[G[v][i]])dfs(G[v][i]);
    }vs.push_back(v);
}
void rdfs(int v,int k){
    used[v]=1;
    cmp[v]=k;
    for(int i=0;i<rG[v].size();i++){
        if(!used[rG[v][i]])rdfs(rG[v][i],k);
    }
}
const int MAX_N=1<<10;
const int MAX_M=1<<11;
int N,M,x,y;
int door[MAX_M][2],key[MAX_N];
int scc(){
    memset(used,0,sizeof(used));
    vs.clear();
    for(int v=0;v<V;v++){if(!used[v])dfs(v);}
    memset(used,0,sizeof(used));
    int k=0;
    for(int i=vs.size()-1;i>=0;i--){
        if(!used[vs[i]])rdfs(vs[i],k++);
    }return k;
}
bool check(int x){
    V=2*N;
    for(int i=0;i<V;i++){G[i].clear();rG[i].clear();}
    for(int i=0;i<x;i++){
        add_edge(key[door[i][0]],door[i][1]);
        add_edge(key[door[i][1]],door[i][0]);
    }scc();
    for(int i=0;i<V;i++){
        if(cmp[i]==cmp[key[i]])return 0;
    }return 1;
}
int solve(){
    int l=0,r=M,ans=0;
    while(l<=r){
        int mid=(l+r)>>1;
        if(check(mid))l=mid+1,ans=mid;
        else r=mid-1;
    }return ans;
}
int main(){
    while(~scanf("%d%d",&N,&M),N+M){
        for(int i=0;i<N;i++){
            scanf("%d%d",&x,&y);
            key[x]=y; key[y]=x; 
        }for(int i=0;i<M;i++)scanf("%d%d",door[i],door[i]+1);
        printf("%d\n",solve());
    }return 0;
}