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.
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
Source
大意：有一只青蛙要跳到另一只青蛙那里问最近的距离是多少，第一个青蛙在1处，第2个青蛙在2处，问你从1跳到2的路径里面最大的所需要的长度。
floyed算法用来算形成的最小正权环，但是也可以用来算从该点出发，到下面点的最小的距离，即把环拉直。
floyed就是三个for循环，遍历每一种的情况，能输出的比较多，优于ballen_ford。
#include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; int const MAX = 300,inf = 0x3f3f3f3f; double map[MAX][MAX]; struct edge{ double x,y; }edge[MAX]; int main(){ int n, count = 1;; while(~scanf("%d",&n)&&n){ for(int i = 1;i <= n ;i++) scanf("%lf%lf",&edge[i].x,&edge[i].y); for(int i = 1; i <= n; i++){ for(int j = i+1; j <= n ;j++){ map[i][j] = sqrt((double)(edge[j].x-edge[i].x)*(edge[j].x-edge[i].x)+(edge[j].y-edge[i].y)*(edge[j].y-edge[i].y)); map[j][i] = sqrt((double)(edge[j].x-edge[i].x)*(edge[j].x-edge[i].x)+(edge[j].y-edge[i].y)*(edge[j].y-edge[i].y)); } } for(int k = 1; k <= n ; k++){ for(int i = 1; i <= n ; i++){ for(int j = 1; j <= n ;j++){ if(map[i][j] > max(map[i][k],map[k][j])) map[i][j] = max(map[i][k],map[k][j]);//因为能往回走 } } } printf("Scenario #%d ",count); printf("Frog Distance = %.3lf ",map[1][2]); count++; } return 0; }