POJ2253Frogger（最短路）

Frogger

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 48889 Accepted: 15576

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

20 03 4317 419 418 50

Sample Output

Scenario #1Frog Distance = 5.000Scenario #2
Frog Distance = 1.414
题意：从第一个坐标到达第二个坐标的最长路径中的最小单步距离。
#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;struct stone{    double x,y;}s[205];int n;double dis[205][205];double dist(stone a,stone b){    return sqrt((a.x-b.x)*(a.x-b.x)              +(a.y-b.y)*(a.y-b.y));}int main(){    int cnt=0;    while(scanf("%d",&n)!=EOF&&n!=0)    {        for(int i=1;i<=n;i++)            scanf("%lf %lf",&s[i].x,&s[i].y);        memset(dis,0,sizeof(dis));        int num=0;        for(int i=1;i<=n;i++)        for(int j=1;j<i;j++)            dis[i][j]=dis[j][i]=dist(s[i],s[j]);       /* for(int i=1;i<=n;i++)            for(int j=1;j<=n;j++)            printf("dis[%d][%d]=%lf\n",i,j,dis[i][j]);*/        for(int k=1;k<=n;k++)            for(int i=1;i<=n;i++)                for(int j=1;j<=n;j++)                dis[i][j]=min(max(dis[i][k],dis[k][j]),dis[i][j]);//从k出发，有两个选择i，j，要使路径最大，选k到另一块石头的最大值，所以是max，//选出来之后k到另一块石头就多了一个跳跃单步距离，所以要比较选最小的跳跃单步距离，//所以max外是min//dis下标用i，j来计数，i，j循环连续，标记二维数组不容易乱        cnt++;        printf("Scenario #%d\n",cnt);        printf("Frog Distance = %.3lf\n\n",dis[1][2]);    }    return 0;}