【tarjan割桥】HDU 4738 Caocao‘s Bridges

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本文探讨了一种算法解决方案，通过Tarjan算法寻找关键桥梁，以帮助周瑜在三国时期的军事冲突中制定策略。该算法确定了切断曹操军队部署的最佳方案。

Problem Description

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.

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

Source

2013 ACM/ICPC Asia Regional Hangzhou Online

tarjan求割桥，然后算每次的最小值

注意：1.如果初始就是不联通，输出0

2.如果最小代价为0，也要改成1，因为要派一个人去

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn=1005;
const int inf=0x3f3f3f3f;
const int maxm=10000005;
int n,m,cnt,times,minn,f[maxn],head[maxn];
int low[maxn],dfn[maxn];
struct edge
{
	int to,nxt,v;
}e[maxm<<1];
void add(int x,int y,int z)
{
	e[++cnt].to=y;
	e[cnt].nxt=head[x];
	e[cnt].v=z;
	head[x]=cnt;
}
void tarjan(int u,int fa)
{
	low[u]=dfn[u]=++times;
	f[u]=fa; int flag=0;
	for(int i=head[u];i;i=e[i].nxt)
	{
		int to=e[i].to;
		if(!flag && to==fa)
		{
			flag=1;
			continue;
		}
		if(!dfn[to])
		{
			tarjan(to,u);
			low[u]=min(low[u],low[to]);
			if(low[to]>dfn[u])
				minn=min(minn,e[i].v);
		}
		else low[u]=min(low[u],dfn[to]);
	}
}
int main()
{
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
	int x,y,z;
	while(scanf("%d%d",&n,&m)==2 && (n || m))
	{
		memset(head,0,sizeof(head)); cnt=0; times=0; minn=inf;
		memset(dfn,0,sizeof(dfn));
		memset(f,0,sizeof(f));
		memset(low,0,sizeof(low));
		for(int i=1;i<=m;i++)
			scanf("%d%d%d",&x,&y,&z),add(x,y,z),add(y,x,z);
		int cc=0;
		for(int i=1;i<=n;i++)
			if(!dfn[i])
			{
				cc++;
				tarjan(i,-1);
			}
		if(cc>1)
		{
			printf("0\n");
			continue;
		}
		if(minn==inf) minn=-1;
		if(!minn) minn=1;
		printf("%d\n",minn);
	}
	return 0;
}