hdu 4738 边双连通&桥，有重边情况

Caocao's Bridges

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1955    Accepted Submission(s): 699

Problem Description

Caocao was defeated by Zhuge Liang and Zhou Yu in the battle of Chibi. But he wouldn't give up. Caocao's army still was not good at water battles, so he came up with another idea. He built many islands in the Changjiang river, and based on those islands, Caocao's army could easily attack Zhou Yu's troop. Caocao also built bridges connecting islands. If all islands were connected by bridges, Caocao's army could be deployed very conveniently among those islands. Zhou Yu couldn't stand with that, so he wanted to destroy some Caocao's bridges so one or more islands would be seperated from other islands. But Zhou Yu had only one bomb which was left by Zhuge Liang, so he could only destroy one bridge. Zhou Yu must send someone carrying the bomb to destroy the bridge. There might be guards on bridges. The soldier number of the bombing team couldn't be less than the guard number of a bridge, or the mission would fail. Please figure out as least how many soldiers Zhou Yu have to sent to complete the island seperating mission.

 

Input

There are no more than 12 test cases.

In each test case:

The first line contains two integers, N and M, meaning that there are N islands and M bridges. All the islands are numbered from 1 to N. ( 2 <= N <= 1000, 0 < M <= N ² )

Next M lines describes M bridges. Each line contains three integers U,V and W, meaning that there is a bridge connecting island U and island V, and there are W guards on that bridge. ( U ≠ V and 0 <= W <= 10,000 )

The input ends with N = 0 and M = 0.

 

Output

For each test case, print the minimum soldier number Zhou Yu had to send to complete the mission. If Zhou Yu couldn't succeed any way, print -1 instead.

 

Sample Input

  

   3 3
1 2 7
2 3 4
3 1 4
3 2
1 2 7
2 3 4
0 0
  

 

Sample Output

  

   -1
4
  

 

Source

2013 ACM/ICPC Asia Regional Hangzhou Online

 

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坑点较多，写在注释里了

#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <stack>
#include <algorithm>
using namespace std;

#define maxn 1005
const int inf =0x3f3f3f3f;
struct edge
{
    int v, w,rev, flag;  //flag用于判断该边是否被访问过，rev代表反向边
    edge(){};
    edge(int vv, int ww, int rr){ v=vv; w=ww;rev=rr; flag=false;}
};
vector<edge> g[maxn];
int low[maxn],dfn[maxn], in[maxn];
int belong[maxn];
stack<int> s;
int n,m, dcc;

void init()
{
    for(int i=0; i<maxn; i++)g[i].clear();
    memset(dfn, 0, sizeof(dfn));
    memset(in, 0, sizeof(in));
    while(!s.empty()) s.pop();
    dcc=0;
}
void add(int u, int v, int w)
{
    g[u].push_back(edge(v, w, g[v].size()));
    g[v].push_back(edge(u, w, g[u].size()-1));
}


void tarjan(int u, int &cnt, int &res)
{
    dfn[u]=low[u]=++cnt;
    s.push(u); in[u]=1;
    int v;
    for(int i=0; i<g[u].size(); i++){  //并不能通过检查v是否为u的父节点而跳过当前循环
        v=g[u][i].v;                   //因为u,v间可能存在多条边，所以需要判断的是是否由同一条边访问回父节点，即同一条边不能来回两次遍历
        if(!dfn[v]){
            g[u][i].flag=true; g[v][g[u][i].rev].flag=true; //标记边访问过，防止同一条边的反向边再次访问
            tarjan(v, cnt, res);
            low[u]=min(low[v], low[u]);
            if(low[v]>dfn[u]){
                res=min(res, g[u][i].w);
            }
        }
        else if(in[v] && !g[u][i].flag) {
            low[u]=min(dfn[v], low[u]);
             g[u][i].flag=true; g[v][g[u][i].rev].flag=true;
        }
    }

    if(low[u]==dfn[u]){
        dcc++;
        do{
            v=s.top(); s.pop();
            belong[v]=dcc;
            in[v]=false;
        }while(v!=u);
    }

}



int main()
{
    while(scanf("%d%d", &n, &m)==2 && n+m!=0){
        init();

        int u,v,w;
        for(int i=0; i<m; i++){
            scanf("%d%d%d", &u, &v, &w);
            add(u,v,w);
        }

        int tot=0, ans=inf;
        tarjan(1, tot, ans);
        if(ans==0) ans=1;  //最小值为0也至少要派1个人去
        for(int i=2; i<=n; i++) //原图不连通，ans=0
            if(!dfn[i]){
                ans=0;
                break;
            }


        if(inf==ans) ans=-1;
        printf("%d\n", ans);
    }
    return 0;
}

。