POJ

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 #2Frog Distance = 1.414

【题意】给你很多只青蛙，给出他们的坐标，问你第一只和第二只的最短距离，最短距离的定义如下：某一条路径的某一次移动的最大值（两点之间的距离）表示该条路径的距离，所有路径的最小距离为最短距离。以第二组测试用例为例：
1->2的距离是2，因为1->2最长的是2
1->3->2的距离是1.414，因为1->3和3->2中最长的是1.414
所以1和2的最短距离是1.414

【分析】题目其实不难，就是不好想，因为点的个数到了200，不知道1是直接到2，还是经过一个点？两个点······后来大佬说是最短路问题，每次更新一条路，直到不能更新为止，一定可以得到最短路径。

【代码】

#include <cstdio>#include <cstring>#include <cmath>#include <queue>using namespace std;struct P{    int x;    int y;};double cal(P a,P b)//用来计算两点之间的距离{    return sqrt((double)((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)));}P poi[300];double mat[222][222];//用来表示两个点之间的距离 int a;void inti(void){    bool change=true;    for(int i=0; i<a; i++)        for(int j=0; j<a; j++)            mat[j][i]=mat[i][j]=cal(poi[i],poi[j]);//先将最长的距离统计出来    while(change)    {        change=false;        for(int i=0; i<a; i++)            for(int j=0; j<a; j++)//不断更新mat[i][j]                 for(int k=0; k<a; k++)                    if(mat[i][j]>max(mat[i][k],mat[k][j]))                    {                        mat[j][i]=mat[i][j]=max(mat[i][k],mat[k][j]);                        change=true;//如果有更新说明还有路不是最短的                    }    }}int main(){    //freopen("in.txt","r",stdin);    int flag=1;    while(scanf("%d",&a)&&a)    {        printf("Scenario #%d\n",flag++);        printf("Frog Distance = ");        for(int i=0; i<a; i++)            scanf("%d %d",&poi[i].x,&poi[i].y);        inti();        printf("%.3f\n\n",mat[0][1]);//被坑了一次，这里说了要有两个回车，哪怕是最后一组用例    }    return 0;}