HDU-4738 Caocao's Bridges

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.

InputThere 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.OutputFor 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

// 无向图找桥

#include <cstdio>
#include <cstring>
#include <algorithm>

#define MAX 1010
#define INF 0x3fffffff

using namespace std;

int n, m;
int tot;
int head[MAX];

typedef struct Edge {
    int to;
    int next;
    int w;
} Edge;
Edge edges[MAX * MAX * 2];

void add_edge( int u, int v, int w ) {
    edges[tot] = (Edge){ v, head[u], w };
    head[u] = tot++;
    edges[tot] = (Edge){ u, head[v], w };
    head[v] = tot++;
}

int DFN[MAX];
int LOW[MAX];
int dfs_clock = 1;
int ans;

void tarjan( int u, int id ) {
    LOW[u] = DFN[u] = dfs_clock++;
    printf( "%d %d\n", u, id );
    for( int e = head[u]; e != -1; e = edges[e].next ) {
        int v = edges[e].to;
        if( e == ( id^1 ) ) continue;   // 不太理解这句话的意思
        if( !DFN[v] ) {
            tarjan( v, e );
            LOW[u] = min( LOW[u], LOW[v] );
            if( LOW[v] > DFN[u] )
                ans = min( ans, edges[e].w );
        }
        else LOW[u] = min( LOW[u], DFN[v] );
    }
}

int main() {
    while( scanf( "%d%d", &n, &m ) != EOF && n ) {
        ans = INF;
        tot = 0;
        dfs_clock = 1;
        memset( DFN, 0, sizeof( DFN ) );
        memset( head, -1, sizeof( head ) );
        for( int i = 0; i < m; i++ ) {
            int u, v, w;
            scanf( "%d%d%d", &u, &v, &w );
            add_edge( u, v, w );
        }

        tarjan( 1, -1 );
        int cnt = 0;
        for( int i = 1; i <= n; i++ ) {
            if( DFN[i] > 0 ) cnt++;
        }
        if( cnt != n ) printf( "0\n" ); //图不连通,不用炸
        else if( ans == INF ) printf( "-1\n" );   //图中无桥
        else if( ans == 0 ) printf( "%d\n", 1 );       //桥上兵为0
        else printf( "%d\n",ans );
    }

    return 0;
}