POJ 2253 Frogger（Dijkstra变形——最短路径最大权值）

题目链接：

http://poj.org/problem?id=2253

Description

Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to visit her, but since the water is dirty and full of tourists' sunscreen, he wants to avoid swimming and instead reach her by jumping.
Unfortunately Fiona's stone is out of his jump range. Therefore Freddy considers to use other stones as intermediate stops and reach her by a sequence of several small jumps.
To execute a given sequence of jumps, a frog's jump range obviously must be at least as long as the longest jump occuring in the sequence.
The frog distance (humans also call it minimax distance) between two stones therefore is defined as the minimum necessary jump range over all possible paths between the two stones.

You are given the coordinates of Freddy's stone, Fiona's stone and all other stones in the lake. Your job is to compute the frog distance between Freddy's and Fiona's stone.

Input

The input will contain one or more test cases. The first line of each test case will contain the number of stones n (2<=n<=200). The next n lines each contain two integers xi,yi (0 <= xi,yi <= 1000) representing the coordinates of stone #i. Stone #1 is Freddy's stone, stone #2 is Fiona's stone, the other n-2 stones are unoccupied. There's a blank line following each test case. Input is terminated by a value of zero (0) for n.

Output

For each test case, print a line saying "Scenario #x" and a line saying "Frog Distance = y" where x is replaced by the test case number (they are numbered from 1) and y is replaced by the appropriate real number, printed to three decimals. Put a blank line after each test case, even after the last one.

Sample Input

2
0 0
3 4

3
17 4
19 4
18 5

0

Sample Output

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

题意描述：
输入几个顶点的坐标
计算并输出从一号顶点到二号顶点所有最短路径中的最长距离
解题思路：
首先求出最短路径，Dijkstra算法不变，变化的是dis数组中存储的是所走的最短路径中最短的一段距离，在更新dis数组是加上判断条件即可。
属于最短路径最大权值，题目很经典，另还需了解
最短路径双重最小权值请参考：http://www.cnblogs.com/wenzhixin/p/7405802.html
最长路径最小权值请参考：http://www.cnblogs.com/wenzhixin/p/7336948.html
AC代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
struct Node
{
    double x,y;
};
struct Node node[210];
double e[210][210],dis[210];
int main()
{
    int n,book[210],i,j,u,v,t=1;
    double x,y,inf=99999999,mins;
    while(scanf("%d",&n),n != 0)
    {
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
                e[i][j]=inf;
        for(i=1;i<=n;i++)
            scanf("%lf%lf",&node[i].x,&node[i].y);
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                if(i != j)
                e[i][j]=sqrt(fabs(node[i].x-node[j].x)*fabs(node[i].x-node[j].x)
                +fabs(node[i].y-node[j].y)*fabs(node[i].y-node[j].y));
            }
        }
        
        for(i=1;i<=n;i++)
            dis[i]=e[1][i];
        memset(book,0,sizeof(book));
        book[1]=1;
        for(i=1;i<=n-1;i++)
        {
            mins=inf;
            for(j=1;j<=n;j++)
            {
                if(!book[j] && dis[j] < mins)
                {
                    mins=dis[j];
                    u=j;
                }
            }
            book[u]=1;
            for(v=1;v<=n;v++)
            {
                if(!book[v] && e[u][v] < inf)
                {
                    if(dis[v] > max(dis[u],e[u][v]))
                    dis[v]=max(dis[u],e[u][v]);    
                }
            }
        }
        printf("Scenario #%d
Frog Distance = %.3lf

",t++,dis[2]);
    }
    return 0;
 } 

 使用Floyd算法更为精简

#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;

const int inf = 99999999;
const int maxn = 210;
struct Node {
    double x,y;
}node[210];

double e[maxn][maxn];

double dis(Node a, Node b) {
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

int main()
{
    int n,kase = 1;
    while(scanf("%d", &n) == 1 && n) {
        for(int i = 1; i <= n ; i++) {
            scanf("%lf%lf", &node[i].x, &node[i].y);
        }        
        
        for(int i = 1; i <=n; i++) {
            for(int j = 1; j <= n; j++){
                e[i][j] = i == j? 0 : inf;
            }
        }
        for(int i = 1; i < n; i++) {
            for(int j = i + 1; j <= n; j++){
                 e[i][j] = e[j][i] = dis(node[i], node[j]);
            }
        }
        
        for(int k = 1; k <= n; k++) {
            for(int i = 1; i <= n; i++) {
                for(int j = 1; j <= n; j++) {
                    e[i][j]=min(e[i][j], max(e[i][k],e[k][j]));
                }
            }
        }
        printf("Scenario #%d
Frog Distance = %.3f

", kase++, e[1][2]);
    }
    return 0;
}