HDU 4738 双连通模版题

九野的博客，转载请注明出处：http://blog.csdn.net/acmmmm/article/details/11711577

题意：给定n个点，m条无向边

下面m行表示u , v ,边权值

求所有桥中最小的桥的权值，如不存在输出-1

若图一开始就不连通或最小权值为0则输出1

双连通求桥裸题

附赠一大波测试数据：

#include <string.h>
#include <cstdio>
#include <vector>
#define N 1010
#define inf 10000000
using namespace std;
inline int Min(int a,int b){return a>b?b:a;}
vector<int>G[N];
int dis[N][N],bian[N][N];
int n,m,f[N];
int find(int x){return x==f[x]?x:f[x]=find(f[x]);}
void Union(int u,int v){
    int fa=find(u),fb=find(v);
    f[fa]=fb;
}

struct node{
    int u,v,d;
}edge[N];//割边不会超过n条
int edgenum;
void PUT(int u,int v,int d){
    node E={u,v,d};
    edge[edgenum++]=E;
}
int pre[N],low[N],dfs_clock;
bool su;
int dfs(int u,int fa){//是连通图,dfs(u, )目的是寻找u的后代所能连回的（最早的祖先）的pre值
    if(su)return 0;
	int lowu=pre[u]= ++ dfs_clock;
    int child=0;
    for(int i=0;i<G[u].size();i++){
        int v=G[u][i];
        if(!pre[v])
		{
            child++;
            int lowv=dfs(v,u);
            lowu = Min(lowu, lowv);
			if(lowv > pre[u]&&bian[u][v]==1){PUT(u,v,dis[u][v]);if(dis[u][v]<=1)return 0;}
        }
        else if(pre[v] < pre[u] && v!=fa)
            lowu = Min(lowu, pre[v]);
    }
    return low[u]=lowu;
}
void findcut(){
    memset(pre,0,sizeof(pre));
    dfs_clock=0;
    dfs(1,-1);//dfs(树根，树根的父亲是-1)
}
int main(){
    int i,u,v,d;
    while(scanf("%d %d",&n,&m),n)
    {
        memset(bian,0,sizeof(bian));
        for(i=1;i<=n;i++)
        {
            G[i].clear();
            f[i]=i;
            for(int j=1;j<=n;j++)
                dis[i][j]=inf;
        }

        while(m--)
        {
            scanf("%d %d %d",&u,&v,&d);
			if(dis[u][v]==inf)
			{
            G[u].push_back(v);       G[v].push_back(u);
            Union(u,v);
			}
            dis[u][v]=dis[v][u]=Min(dis[u][v],d);
            bian[u][v]++;            bian[v][u]++;
        }

        bool fa=false;	find(1);
        for(i=1;i<=n;i++)if(find(i)!=f[1]){fa=true;break;}
        if(fa){printf("0\n");continue;}

		su=false;
		edgenum=0;
        findcut();

        if(edgenum)
        {
            int ans=inf;
            for(i=0;i<edgenum;i++)ans=Min(ans,edge[i].d);
            if(ans==inf)ans=-1;
            else if(ans==0)ans=1;
            printf("%d\n",ans);
        }
        else printf("-1\n");
    }
    return 0;
}
/*
3 3
1 2 7
2 3 4
3 1 4 

3 2
1 2 7
2 3 4 

3 4
1 2 7
2 1 7
2 3 4
3 2 4 

4 3
1 2 1
1 2 3
3 4 4 

4 4
1 2 1
1 2 3
3 4 4
2 4 6 

4 5
1 2 1
1 2 3
3 4 4
2 4 6
1 3 0 

4 5
4 3 0
3 4 4
2 4 6
2 3 0
1 3 0 

2 1
1 2 0 

4 2
1 2 3
1 3 5 

4 5
1 2 3
2 3 4
1 3 5
4 3 7
4 3 6 

4 4
1 2 3
2 3 4
1 3 5
4 3 6 

4 4
1 2 4
1 3 5
4 3 6
3 4 7 

6 7
1 2 1
1 3 2
2 3 3
3 4 4
4 6 5
4 5 6
5 6 7 

5 6
1 2 3
1 3 4
2 3 5
3 4 7
3 5 8
5 4 9 

8 9
1 8 1
5 1 2
1 7 3
1 4 4
1 2 5
4 3 6
3 2 7
5 6 8
6 7 9 

6 6
1 4 1
1 2 4
2 5 4
5 6 4
3 6 4
4 3 2 

6 5
1 2 4
2 5 4
5 6 4
3 6 4
4 3 2 

8 10
1 8 1
5 1 2
1 7 3
1 4 4
1 2 5
4 3 6
3 2 7
5 6 8
6 7 9
8 1 2 

3 3
1 2 7
1 3 1
1 3 4 

0 0
*/