hdu4738（双连通分量）

最新推荐文章于 2021-10-14 21:04:13 发布

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最新推荐文章于 2021-10-14 21:04:13 发布

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分类专栏：图论_双连通分量文章标签： ACM hdu

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本文解析了一道经典的算法题目“曹操之桥”，探讨如何在无向图中找到关键的桥，即一旦被移除会导致图分裂的边，并确定完成这一任务所需的最少兵力。

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Caocao's Bridges

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1499 Accepted Submission(s): 566

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

Sample Output

-1
4

题意：给一个无向图，求找出一条边，使得删除这条边以后整个图就不连通了，找出满足的边里权值最小的那个，找不到输出-1

思路：这题一看就是无向图中找桥的问题，直接套板子，但是这题有两个超大的坑

1.如果最开始图就不连通，直接输出0

2.如果找到的桥的最小权值为0，则不能输出0，而要输出1（即使没有守卫也至少要派一个人去炸吧，桥不可能自己炸吧，又不是有特异功能？）

#include 
#include 
#include 
#include 
using namespace std;
#define maxn 1010
#define INF (~0U>>1)

int ans;
int head[1010];
int edge[1010*1010*2];
int next[1010*1010*2];
int ww[1010*1010*2];
int d;
int n,m;
int low[maxn];
int pre[maxn];
int dfs_clock;
int vis[maxn];
int vv[maxn][maxn];

int mini(int x,int y)
{
    return xpre[u]){
                if(vv[u][v]==1&&ans>ww[i])ans=ww[i];
            }
        }
        else if(pre[v]w)temp=w;
            vv[u][v]++;
            vv[v][u]++;
            add(u,v,w);
            add(v,u,w);
        }
        memset(vis,0,sizeof(vis));
        ddf(1);
        for(i=1;i<=n;i++)if(!vis[i])break;
        int ok=0;
        if(i<=n)ok=1;

        if(ok==1)printf("0\n");
        else {
        dfs_clock=0;
        memset(pre,0,sizeof(pre));
        memset(low,0,sizeof(low));
        ans=INF;
        dfs(1,0);


        if(ans==INF)printf("-1\n");
        else {

        if(ans==0)printf("%d\n",1);
        else printf("%d\n",ans);
        }


        }
    }
}