hdu 4387 双连通分量求权值最小的桥

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探讨在曹操通过建造岛屿和桥梁增强军队部署灵活性的情况下，周瑜如何利用有限资源切断曹操军队的连接。需找出最少士兵数量完成拆桥任务。

Caocao's Bridges

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3639 Accepted Submission(s): 1146

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

Sample Output

  
   -1
4

Source

2013 ACM/ICPC Asia Regional Hangzhou Online

解析：
求无向图的双连通分量的桥，若本来图就没有连通，就不派人去，若桥的权值为0，还要派1个人去。求桥用tarjan算法。
代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn = 1000000+10;
const int Maxn = 1000+10;
const int inf = 0x3f3f3f3f;
int n, m;
struct EDGE
{
    int t, c, next;
}edge[maxn];

int dfn[Maxn], low[Maxn];
vector<int>ans[10000+10];

int g[Maxn][Maxn], head[Maxn];

int cnt = 0;

void addedge(int u, int v, int c) {
    edge[cnt].t = v;
    edge[cnt].c = c;
    edge[cnt].next = head[u];
    head[u] = cnt;
    cnt++;
    edge[cnt].t = u;
    edge[cnt].c = c;
    edge[cnt].next = head[v];
    head[v] = cnt;
    cnt++;
}


int stack[Maxn], sk;
void tarjan(int now, int &sig, int &num, int from) {
    dfn[now] = low[now] = ++sig;
    for (int i=head[now]; i!=-1; i=edge[i].next) {
        if (i == (from^1))
            continue;
        if (!dfn[edge[i].t]) {
            tarjan(edge[i].t, sig, num, i);
            low[now] = min(low[now], low[edge[i].t]);
            if (low[edge[i].t] > dfn[now]) {
                ans[num].push_back(now);
                ans[num].push_back(edge[i].t);
                num++;
            }
        }
        else            
           low[now] = min(low[now], dfn[edge[i].t]);
    }
}
int vis[Maxn];
void dfs(int u, int &len) {
    vis[u] = 1;
    for (int i=head[u]; i!=-1; i=edge[i].next)
        if (!vis[edge[i].t]) 
            len++, dfs(edge[i].t, len);
}


int main()
{
    while (scanf("%d%d", &n, &m)!=EOF) {
        if (n==0 && m==0) break;
        cnt = 0;
        memset(low, 0, sizeof(low));
        memset(dfn, 0, sizeof(dfn));
        memset(g, inf, sizeof(g));;
        memset(head, -1, sizeof(head));
        for (int i=0; i<m; i++) {
            int u, v, w;
            scanf("%d%d%d", &u, &v, &w);
            g[u][v] = w;
            g[v][u] = w;
            addedge(u, v, w);
        }
        int sig, num, res = inf;
        sig = num = 0;
        memset(vis, 0, sizeof(vis));
        int len=1;
        dfs(1, len);
        if (len < n) {
            printf("0\n");
            continue;
        }
	    tarjan(1, sig, num, -1);
        for (int i=0; i<num; i++) {
            res = min(res, g[ans[i][0]][ans[i][1]]);
        }
        for (int i=0; i<num; i++)
            ans[i].clear();

        if (res != inf && res != 0)
            printf("%d\n", res);
        else if (res == 0)
            cout << "1" << endl;
        else
            cout << "-1" << endl;
    }
	return 0;
}