HDU 4738 Caocao's Bridges (Bridge, any bit operation must be bracketed, because there are heavy edges so use forward star)

HDU 4738 Caocao's Bridges (Bridge, any bit operation must be bracketed, because there are heavy edges so use forward star)

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.

Question: Cao Cao has an undirected graph with some edges in the picture. Zhou Yu now wants to send some people to blow up one edge so that Cao Cao's undirected graph is divided into different parts. Each edge of Cao Cao has guards. Zhou Yu sent The number of people cannot be less than the number of guards on the side. Ask how many people should be sent at least.

• Pay attention to heavy side processing, use chain forward star.
• When the guard is 0, you also need to send a person;
• If the original picture is not connected (tarjan multiple times), there is no need to send someone, the answer is 0.
• The answer directly outputs the smallest edge weight among all bridges.

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
const int N = 500007, M = 5000007, INF = 0x3f3f3f3f;

int n, m;
int bridge[N];
int dfn[N], low[N], num;
int head[N], ver[M], nex[M], edge[M], tot;
int ans;
int dcc_cnt;
int min_bridge;

void init()
{
    
    
    num = tot = dcc_cnt = 0;
    memset(head, -1, sizeof head);
    memset(dfn, 0, sizeof dfn);
    memset(low, 0, sizeof low);
    min_bridge = INF;
}

void add(int x, int y, int z)
{
    
    
    ver[tot] = y;
    edge[tot] = z;
    nex[tot] = head[x];
    head[x] = tot ++ ;
}

void tarjan(int x, int in_edge)
{
    
    
    dfn[x] = low[x] = ++ num;
    for(int i = head[x] ;~i; i = nex[i]){
    
    
        int y = ver[i], z = edge[i];
        if(!dfn[y]){
    
    
            tarjan(y, i);
            low[x] = min(low[x], low[y]);
            if(dfn[x] < low[y]){
    
    
                bridge[i] = bridge[i ^ 1] = true;
                min_bridge = min(min_bridge, z);
            }
        }
        else if(i != (in_edge ^ 1))
            low[x] = min(low[x], dfn[y]);
    }
}

int main()
{
    
    
    while(~scanf("%d%d", &n, &m) && n ){
    
    
        init();

        for(int i = 1; i <= m; ++ i){
    
    
            int x, y ,z;
            scanf("%d%d%d", &x, &y, &z);
            add(x, y, z), add(y, x, z);
        }
        dcc_cnt = 0;
        for(int i = 1; i <= n; ++ i)
            if(!dfn[i])
                tarjan(i, 0), dcc_cnt ++ ;
        
        if(dcc_cnt > 1){
    
    //tarjan多次说明不连通，直接输出0
            printf("0\n");
            continue;
        }
        if(min_bridge == INF)min_bridge = -1;
        else if(min_bridge == 0)min_bridge = 1;
        printf("%d\n", min_bridge);
    }
    return 0;
}