Caocao's Bridges

Caocao's Bridges

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2616    Accepted Submission(s): 849

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 <= N² )

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

这题的意思就是求出所有的桥，然后输出桥的权值的最小值。

但是坑点比较多。

如果一开始是不连通的，输出0.

图有重边，需要处理。

还有如果取到的最小值是0的话，要输出1，表示要派一个人过去。

#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <limits>
#include <stack>
#include <vector>
#include <map>

using namespace std;

#define N 1350
#define INF 0xfffffff
#define PI acos (-1.0)
#define EPS 1e-83

struct node
{
    int v, next, flow;
}edge[N*N];

int n, m, Time, minx, cnt;
int low[N], dfn[N], f[N], head[N];

vector<vector<int> > G;

void init()
{
    cnt = Time = 0;
    memset(low, 0, sizeof(low));
    memset(dfn, 0, sizeof(dfn));
    memset(f, 0, sizeof(f));
    memset(head, -1, sizeof(head));
}

void addedge(int u, int v, int flow)   // 邻接表存边
{
    edge[cnt].v = v;
    edge[cnt].flow = flow;
    edge[cnt].next = head[u];
    head[u] = cnt;
    cnt++;
}

void Tarjan(int u, int fa)
{
    f[u] = fa;
    low[u] = dfn[u] = ++Time;
    int k = 0;     // 去重边

    for(int i = head[u]; i != -1; i = edge[i].next)
    {
        int v = edge[i].v;
        if(fa == v && !k)
        {
            k++;
            continue;
        }
        if(!low[v])
        {
            Tarjan(v, u);
            low[u] = min(low[u], low[v]);
            if(low[v] > dfn[u])
                minx = min(minx, edge[i].flow);  // 求桥的最小值
        }
        else
            low[u] = min(low[u], dfn[v]);
    }
}

int main()
{
    int u, v, flow, k;

    while(scanf("%d%d", &n, &m), n+m)
    {
        init();

        while(m--)
        {
            scanf("%d%d%d", &u, &v, &flow);
            addedge(u, v, flow);
            addedge(v, u, flow);
        }
        k = 0, minx = INF;

        for(int i = 1; i <= n; i++)
        {
            if(!low[i])
            {
                Tarjan(i, -1);
                k++;   // 判断有几个连通图，
            }
        }
        if(k > 1) // 如果连通图大于1，输出0
        {
            puts("0");
            continue;
        }
        if(!minx)  // 桥最小值是0，需要派一个区攻击
            minx++;
        else if(minx == INF)  // 没有桥
            minx = -1;
        printf("%d
", minx);
    }
    return 0;
}