hdu 4738 Caocao's Bridges（桥的最小权值+去重）

http://acm.hdu.edu.cn/showproblem.php?pid=4738

题目大意：曹操有一些岛屿被桥连接，每座都有士兵把守，周瑜想把这些岛屿分成两部分，但他只能炸毁一条桥，问最少需要派几个士兵去;如果不能完成输出-1

1：如果这些岛屿不连通，则不需要派人前去

2：如果桥的守卫是0的话也得派一人去炸毁

3：如果不能完成输出-1

4：输出最少需派的人数

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<queue>
#include<algorithm>
using namespace std;
#define N 1100
#define INF 0xfffffff

int dfn[N], low[N], head[N], f[N];
int Time, m, n, Min, cnt;
struct Edge
{
    int next, u, v, c;
} edge[N * N];
void Init()
{
    memset(head, -1, sizeof(head));
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(f, 0, sizeof(f));
    Time = cnt = 0;
}

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

void Tarjan(int u, int fa)
{
    int i, v, flag = 0;
    low[u] = dfn[u] = ++Time;
    f[u] = fa;
    for(i = head[u] ; i != -1 ; i = edge[i].next)
    {
        v = edge[i].v;
        if(v == fa && !flag)
        {
            flag = 1;
            continue;
        }//去重边，如果有重边则跳过
        if(!dfn[v])
        {
            Tarjan(v, u);
            low[u] = min(low[u], low[v]);
            if(dfn[u] < low[v])
                Min = min(Min, edge[i].c);
        }
        else
            low[u] = min(low[u], dfn[v]);
    }
}

int main()
{
    int i, u, v, c, k;
    while(scanf("%d%d", &m, &n), m + n)
    {
        Init();
        k = 0;
        while(n--)
        {
            scanf("%d%d%d", &u, &v, &c);
            AddEdge(u, v, c);
            AddEdge(v, u, c);
        }
        Min = INF;
        for(i = 1 ; i <= m ; i++)
        {
            if(!dfn[i])
            {
                Tarjan(i, -1);
                k++;
            }
        }
        if(k > 1)
        {
            printf("0
");
            continue;
        }//1
        if(Min == 0)
            printf("1
");//2
        else if(Min == INF)
            printf("-1
");//3
        else
            printf("%d
", Min);//4
    }
    return 0;
}