POJ - 2253 Frogger（Floyd最短路+预处理）

 1 #include <cmath>
 2 #include <cstdio>
 3 #include <algorithm>
 4 using namespace std;
 5 
 6 int n,cas=1;
 7 double dp[222][222];
 8 struct point{
 9     double x;
10     double y;
11 }P[222];
12 
13 double fun(point a,point b){
14     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
15 }
16 
17 void Floyd(){
18     for(int k=1;k<=n;k++)
19     for(int i=1;i<=n;i++)
20     for(int j=1;j<=n;j++)
21     if(dp[i][j]>dp[i][k]&&dp[i][j]>dp[k][j])
22     dp[i][j]=max(dp[i][k],dp[k][j]);
23 }
24 
25 int main(){
26     while(scanf("%d",&n)!=EOF&&n){
27         for(int i=1;i<=n;i++) scanf("%lf %lf",&P[i].x,&P[i].y);
28         for(int i=1;i<=n;i++) 
29         for(int j=1;j<=n;j++)
30         dp[i][j]=fun(P[i],P[j]);
31         Floyd();
32         printf("Scenario #%d\nFrog Distance = %.3f\n\n",cas++,dp[1][2]);
33     }
34     return 0;
35 }