HDU 4738 Caocao's Bridges

题目大意：

曹操有很多岛屿，然后呢需要建造一些桥梁将所有的岛屿链接起来，周瑜要做的是就是不让曹操将所有岛屿连接起来，每个座桥有人在守卫， 周瑜只能炸一座桥，并且他派人去炸桥只能派的人数必须 大于等于守桥的人数。

输出最小的炸桥人数， 要是没有答案就输出 -1

 

 

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <vector>
usingnamespace std;
#define INF 0x7ffffff
#define maxn 100005
typedef longlong LL;
#define Min(a,b) (a<b?a:b)
struct node
{
    int e, w;
    node(int e=0,int w=0): e(e), w(w) {}
};
int low[maxn], dfn[maxn], Time, m, n;
int Father[maxn];
int bridge[maxn];
vector<vector<node> > G;
void init()
{
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(Father, 0,sizeof(Father));
    memset(bridge, -1, sizeof(bridge));
    Time = 0;
    G.clear();
    G.resize(n+2);
}

void Tarjan(int u,int fa)
{
    low[u] = dfn[u] = ++Time;
    Father[u] = fa;
    int len = G[u].size(), k = 0;
    node v;
    for(int i=0; i<len; i++)
    {
        v = G[u][i];
        if(!k && v.e == fa)
        {
            k ++;
            continue;
        }
        if(!low[v.e])
        {
            Tarjan(v.e, u);
            low[u] = min(low[u], low[v.e]);
        }
        else
        {
            low[u] = min(low[u], dfn[v.e]);
        }
        if(dfn[u] < low[v.e])
            bridge[v.e] = v.w;

        if(k == 2 && bridge[v.e] != -1)
            bridge[v.e] = -1;
    }
}

void solve()
{
    int ans = INF, i;
    Tarjan(1,1);
    for(i=2; i<=n; i++)
    {
        if( !low[i] )
            break;
    }
    if(i != n+1)
    {
        puts("0");
        return ;
    }

    for(i=1; i<=n; i++)
    {
        if(bridge[i] != -1)
        {
            ans = min(bridge[i], ans);
        }
    }

    if(ans == 0)
        ans ++;
    if(ans == INF)
        ans = -1;
    printf("%d
", ans);
}
int main()
{
    while(scanf("%d %d",&n, &m), m+n)
    {
        init();
        for(int i=0; i<m; i++)
        {
            int a, b, c;
            scanf("%d %d %d",&a, &b, &c);
            G[a].push_back(node(b,c));
            G[b].push_back(node(a,c));
        }
        solve();
    }
    return0;
}