hdu 5137 How Many Maos Does the Guanxi Worth(dijkstra算法+删边)

题目来源：hdu-5137 How Many Maos Does the Guanxi Worth

How Many Maos Does the Guanxi Worth
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 1058 Accepted Submission(s): 373

Problem Description
“Guanxi” is a very important word in Chinese. It kind of means “relationship” or “contact”. Guanxi can be based on friendship, but also can be built on money. So Chinese often say “I don’t have one mao (0.1 RMB) guanxi with you.” or “The guanxi between them is naked money guanxi.” It is said that the Chinese society is a guanxi society, so you can see guanxi plays a very important role in many things.

Here is an example. In many cities in China, the government prohibit the middle school entrance examinations in order to relief studying burden of primary school students. Because there is no clear and strict standard of entrance, someone may make their children enter good middle schools through guanxis. Boss Liu wants to send his kid to a middle school by guanxi this year. So he find out his guanxi net. Boss Liu’s guanxi net consists of N people including Boss Liu and the schoolmaster. In this net, two persons who has a guanxi between them can help each other. Because Boss Liu is a big money(In Chinese English, A “big money” means one who has a lot of money) and has little friends, his guanxi net is a naked money guanxi net – it means that if there is a guanxi between A and B and A helps B, A must get paid. Through his guanxi net, Boss Liu may ask A to help him, then A may ask B for help, and then B may ask C for help …… If the request finally reaches the schoolmaster, Boss Liu’s kid will be accepted by the middle school. Of course, all helpers including the schoolmaster are paid by Boss Liu.

You hate Boss Liu and you want to undermine Boss Liu’s plan. All you can do is to persuade ONE person in Boss Liu’s guanxi net to reject any request. This person can be any one, but can’t be Boss Liu or the schoolmaster. If you can’t make Boss Liu fail, you want Boss Liu to spend as much money as possible. You should figure out that after you have done your best, how much at least must Boss Liu spend to get what he wants. Please note that if you do nothing, Boss Liu will definitely succeed.

Input
There are several test cases.

For each test case:

The first line contains two integers N and M. N means that there are N people in Boss Liu’s guanxi net. They are numbered from 1 to N. Boss Liu is No. 1 and the schoolmaster is No. N. M means that there are M guanxis in Boss Liu’s guanxi net. (3 <=N <= 30, 3 <= M <= 1000)

Then M lines follow. Each line contains three integers A, B and C, meaning that there is a guanxi between A and B, and if A asks B or B asks A for help, the helper will be paid C RMB by Boss Liu.

The input ends with N = 0 and M = 0.

It’s guaranteed that Boss Liu’s request can reach the schoolmaster if you do not try to undermine his plan.

Output
For each test case, output the minimum money Boss Liu has to spend after you have done your best. If Boss Liu will fail to send his kid to the middle school, print “Inf” instead.

Sample Input
4 5
1 2 3
1 3 7
1 4 50
2 3 4
3 4 2
3 2
1 2 30
2 3 10
0 0

Sample Output
50
Inf

题目大意：
给出n个点，m条边，除了点1和点n，可以任意删去一个点，求从1到n的最短路中最长的距离。
题目分析：
首先需要写出一个dijkstara算法求最短路，然后需要进行删点操作，删去一个点，即将与该点相连的所有路均设为无穷大即可，依次枚举删去各个点的最短路，取最大值即为所求值。
AC代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define INF 0x3f3f3f3f
#define MAX 10010
using namespace std;
int map[MAX][MAX];          //用来存图 
int low[MAX][MAX];          //用来记录所删去的边 
int mark[MAX];              //用来标记点是否使用过 
int dis[MAX];               //存放各点距起点的最短路 
int n,m;
int dijkstra(int s)         //dijkstra算法 
{
    memset(mark,0,sizeof(mark));
    for(int i=0;i<=n;i++)
    dis[i]=map[s][i];
    mark[s]=1;
    dis[s]=0;
    for(int i=1;i<n;i++)
    {
        int min=INF,vir=0;
        for(int j=1;j<=n;j++)
        if(!mark[j]&&dis[j]<min)
        {
            min=dis[j];
            vir=j;
        }
        mark[vir]=1;
        for(int j=1;j<=n;j++)
        if(!mark[j]&&dis[j]>dis[vir]+map[vir][j])
        dis[j]=dis[vir]+map[vir][j];
    }   
    return dis[n];                  //返回点1~n的最短距离 
}
int main()
{
    while(scanf("%d%d",&n,&m),n&&m)
    {
        for(int i=0;i<=n;i++)       //初始化 
        for(int j=0;j<=n;j++)
        if(i==j)
            map[i][j]=0;
        else
            map[i][j]=INF;
        for(int i=0;i<m;i++)        //存图，并进行了去重操作 
        {
            int a,b,d;
            scanf("%d%d%d",&a,&b,&d);
            if(map[a][b]>d)
            {
                map[a][b]=d;
                map[b][a]=d;
            }
        }
        int ans=0;                  //用来找最大值 
        for(int i=2;i<n;i++)        //枚举要删去的点，2~n 
        {
            for(int j=1;j<=n;j++)   //将与此点有关的路径置为无穷大 
            {
                low[i][j]=map[i][j];    //low数组暂时存放所删去边的权值，便于恢复 
                map[i][j]=map[j][i]=INF;
            }
            ans=max(ans,dijkstra(1));   //ans记录的是最短路径中最大的值 
            for(int j=1;j<=n;j++)       //恢复边 
            {
                map[i][j]=map[j][i]=low[i][j];
            }
        }
        if(ans==INF)
            printf("Inf\n");
        else
            printf("%d\n",ans);
    }
    return 0;
}