hdu 4738 Caocao's Bridges (tarjan求桥)

• 题意: 求权值最小的割边
• 思路: \(low[v]>dfn[u]\) 则 \(u-v\) 这条边则为割边,由于有重边(重边一定不是割边),加个判断是否为第一次访问到父亲即可.

#include <iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<cstdio>
#define ll long long
using namespace std;
const int N = 1e5+10;
const int M = 1e6+10;
int n,m;
int ans;
struct Edge{
    int fr,to,nxt,w;
}edge[M];
int tot,head[N];
void init(){
    tot = 0;
    ans = 1<<30;
    memset(head,-1,sizeof head);
    memset(dfn,0,sizeof dfn);
    memset(low,0,sizeof low);
}
void add(int u,int v,int w){
    edge[tot].fr = u;
    edge[tot].to = v;
    edge[tot].w = w;
    edge[tot].nxt = head[u];
    head[u] = tot++;
}
int dfn[N],low[N];
int ind;
void tarjan(int u,int fa){
    int v;
    dfn[u] = low[u] = ++ind;
    int k = 0;
    for(int i =head[u];i!=-1;i=edge[i].nxt){
        v = edge[i].to;
        if(v== fa && !k){   // 处理重边
            k++;
            continue;
        }
        if(!dfn[v]){
            tarjan(v,u);
            low[u] = min(low[u],low[v]);
            if(low[v]>dfn[u]){      // 割桥
                ans = min(ans,edge[i].w);
            }
        }
        else {
            low[u] = min(low[u],dfn[v]);
        }
    }
}
void solve(){
    init();
    ind  = 0 ;
    int u,v,w;
    for(int i=1;i<=m;++i){
        scanf("%d %d %d",&u,&v,&w);
        add(u,v,w);
        add(v,u,w);
    }
    tarjan(1,-1);
    for(int i=1;i<=n;++i){
        if(!dfn[i]){
            printf("0\n");
            return ;
        }
    }
    if(ans == (1<<30))  ans = -1;
    if(ans == 0 )   ans++;
    printf("%d\n",ans);
}
int main()
{
    while(scanf("%d%d",&n,&m)){
        if(n==0 && m==0)    break;
        solve();
    }
    return 0;
}

多条割边求最小值,割边不存在则输出-1,割边权为0要输出1,不连通输出0