Caocao's Bridges HDU - 4738 求桥

题目描述

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.

输入格式

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 <= 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.

输出格式

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.

样例

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

分析

题意很简单，就是求图中最小桥的价值

但要注意以下几点

1、如果图不是联通的，输出0

2、如果最小桥的值为0，不能输出0，要输出1，因为你至少要派一名士兵去

3、有可能有重边

代码

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
const int maxn=2000005;
int head[maxn],tot=2;
struct asd{
	int val,to,next;
}b[maxn];
void ad(int aa,int bb,int cc){
	b[tot].to=bb;
	b[tot].val=cc;
	b[tot].next=head[aa];
	head[aa]=tot++;
}
int dfn[maxn],low[maxn],dfnc;
bool bri[maxn];
void tarjan(int now,int id){
	dfn[now]=low[now]=++dfnc;
	for(int i=head[now];i!=-1;i=b[i].next){
		if(i==(id^1)) continue;
		int u=b[i].to;
		if(!dfn[u]){
			tarjan(u,i);
			low[now]=min(low[now],low[u]);
			if(dfn[now]<low[u]){
				bri[i]=bri[i^1]=1;
			}
		} else {
			low[now]=min(low[now],dfn[u]);
		}
	}
}
int main(){
	int n,m;
	while(scanf("%d%d",&n,&m)!=EOF && n!=0){
		memset(head,-1,sizeof(head));
		memset(&b,0,sizeof(struct asd));
		memset(dfn,0,sizeof(dfn));
		memset(low,0,sizeof(low));
		memset(bri,0,sizeof(bri));
		dfnc=0;
		tot=2;
		for(int i=1;i<=m;i++){
			int aa,bb,cc;
			scanf("%d%d%d",&aa,&bb,&cc);
			ad(aa,bb,cc),ad(bb,aa,cc);
		}
		tarjan(1,-1);
		bool jud=0;
		for(int i=1;i<=n;i++){
			if(!dfn[i]){
				jud=1;
			}
		}
		if(jud){
			printf("0\n");
			continue;
		}
		int ans=0x3f3f3f3f;
		for(int i=2;i<tot;i++){
			if(bri[i]==1){
				ans=min(ans,b[i].val);
			}
		}
		if(ans==0x3f3f3f3f) printf("-1\n");
		else printf("%d\n",max(ans,1));
	}
	return 0;
}