LOJ-10102（桥的判断）

题目链接：传送门

思路：找桥就行了，条件是num[v]<low[u],pre!=v;

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<stack>
using namespace std;
const int maxn = 30030;
int num[maxn],vis[maxn],low[maxn],tim,cnt;
vector <int> vc[maxn];
int MIN(int x,int y)
{
    return x<y?x:y;
}
int MAX(int x,int y)
{
    return x>y?x:y;
}
void Init()
{
    memset(vis,0,sizeof(vis));
    memset(num,0,sizeof(num));
    memset(low,0,sizeof(low));
    for(int i=0;i<maxn;i++) vc[i].clear();
    tim=0;cnt=0;
}
void Tarjan(int u,int pre)
{
    num[u]=low[u]=++tim;
    vis[u]=1;
    int v,i;
    for(i=0;i<vc[u].size();i++){
        v=vc[u][i];
        if(pre==v) continue;
        if(!vis[v]){
            Tarjan(v,u);
            low[u]=MIN(low[u],low[v]);
            if(num[u]<low[v]) cnt++;
        }
        else low[u]=MIN(low[u],num[v]);
    }
}
int main(void)
{
    int n,m,i,j,x,y;
    while(~scanf("%d%d",&n,&m)&&(n+m)){
        Init();
        for(i=1;i<=m;i++){
            scanf("%d%d",&x,&y);
            vc[x].push_back(y);
            vc[y].push_back(x);
        }
        Tarjan(1,-1);
        printf("%d\n",cnt);
    }
    return 0;
}

View Code