hdu1595 find the longest of the shortest(Dijkstra)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1595

find the longest of the shortest

Time Limit: 1000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1665    Accepted Submission(s): 588

Problem Description

Marica is very angry with Mirko because he found a new girlfriend and she seeks revenge.Since she doesn't live in the same city, she started preparing for the long journey.We know for every road how many minutes it takes to come from one city to another.
Mirko overheard in the car that one of the roads is under repairs, and that it is blocked, but didn't konw exactly which road. It is possible to come from Marica's city to Mirko's no matter which road is closed.
Marica will travel only by non-blocked roads, and she will travel by shortest route. Mirko wants to know how long will it take for her to get to his city in the worst case, so that he could make sure that his girlfriend is out of town for long enough.Write a program that helps Mirko in finding out what is the longest time in minutes it could take for Marica to come by shortest route by non-blocked roads to his city.

 

Input

Each case there are two numbers in the first row, N and M, separated by a single space, the number of towns,and the number of roads between the towns. 1 ≤ N ≤ 1000, 1 ≤ M ≤ N*(N-1)/2. The cities are markedwith numbers from 1 to N, Mirko is located in city 1, and Marica in city N.
In the next M lines are three numbers A, B and V, separated by commas. 1 ≤ A,B ≤ N, 1 ≤ V ≤ 1000.Those numbers mean that there is a two-way road between cities A and B, and that it is crossable in V minutes.

 

Output

In the first line of the output file write the maximum time in minutes, it could take Marica to come to Mirko.

 

Sample Input

5 6
1 2 4
1 3 3
2 3 1
2 4 4
2 5 7
4 5 1

6 7
1 2 1
2 3 4
3 4 4
4 6 4
1 5 5
2 5 2
5 6 5

5 7
1 2 8
1 4 10
2 3 9
2 4 10
2 5 1
3 4 7
3 5 10

 

Sample Output

11
13
27

 

Author

ailyanlu

 

Source

HDU 2007-Spring Programming Contest - Warm Up （1） 

题意：如果图中某条路径被堵死。它的最坏情况下的最短路径是多少？基本算法就是先求出最短路径。然后如果最短路径中的某一条边被堵死。再求最短路，取这些最短路的最大值就可以。

代码例如以下：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdlib>
#include <climits>
#include <ctype.h>
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;
#define PI acos(-1.0)
#define INF 0x3fffffff
//typedef long long LL;
//typedef __int64 LL;
#define MAXN 1017
int mat[MAXN][MAXN];
int n, m;
int re[MAXN];//记录路径
void init()
{
	for(int i = 0; i <= n; i++)
	{
		re[i] = 0;
		for(int j = 0 ; j <= n; j++)
		{
			if(i == j)
				mat[i][j] = 0;
			else
				mat[i][j] = INF;
		}
	}
}

int dijkstra (int f)
{
	int dis[MAXN];//记录到随意点的最短距离
	int mark[MAXN];//记录被选中的结点 
	int i,j,k = 0;
	for(i = 0 ; i <= n ; i++)//初始化全部结点，每一个结点都没有被选中 
		mark[i] = 0;
    for(i = 0 ; i <= n ; i++)
    {
        dis[i] = INF;
    }
	//	mark[1] = 1;
    dis[1] = 0;//start为1
    int min ;//设置最短的距离。 
    for(i = 1 ; i <= n; i++)
    {
        min = INF;
        for(j = 1 ; j <= n;j++)
        {
            if(mark[j] == 0  && dis[j] < min)//未被选中的结点中，距离最短的被选中 
            {
                min = dis[j] ;
                k = j;
            }
        }
        mark[k] = 1;//标记为被选中 
        for(j = 1 ; j <= n ; j++)
        {
            if( mark[j] == 0  && (dis[j] > (dis[k] + mat[k][j])))//改动剩余结点的最短距离 
            {
                dis[j] = dis[k] + mat[k][j];
				if(f)
					re[j] = k;
            }
        }
    }
    return dis[n];    
} 

int main()
{
	int i, j;
	int a, b, v;
	while(~scanf("%d%d",&n,&m))
	{
		init();
		for(i = 0; i < m; i++)
		{
			scanf("%d%d%d",&a,&b,&v);
			if(v < mat[a][b])
			{
				mat[a][b] = mat[b][a] = v;
			}
		}
		int ans = dijkstra(1);
		for(i = n; i != 1; i = re[i])
		{
			int t = mat[i][re[i]];
			mat[i][re[i]] = INF;
			mat[re[i]][i] = INF;
			int tt = dijkstra(0);//从0開始为了re[]不再记录路径
			if( ans < tt)
				ans = tt;
			mat[i][re[i]] = t;
			mat[re[i]][i] = t;
		}
		printf("%d
",ans);
	}
	return 0;
}