POJ2253 frogger 最短路 floyd

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<ctype.h>
#include<queue>
#include<stack>
#include<stdlib.h>
#include<math.h>
#include<limits.h>
#define max(a, b) a>b?a:b;
#define min(a, b) a<b?a:b;
using namespace std;
const int N = 210;
struct node
{
int x, y;
} a[N];
int n;
double f[N][N];
void floyd()
{
for(int k=1; k<=n; k++)
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
{
double temp=max(f[i][k], f[k][j])
if(temp<f[i][j])
f[i][j]=temp;
}
}
int main()
{

int Case=0;
while(scanf("%d", &n), n)
{
memset(a, 0, sizeof(a));
memset(f, 0, sizeof(f));

for(int i=1; i<=n; i++)
{
scanf("%d%d", &a[i].x, &a[i].y);
}
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
f[i][j]=sqrt(1.0*(a[i].x-a[j].x)*(a[i].x-a[j].x)+1.0*(a[i].y-a[j].y)*(a[i].y-a[j].y));
floyd();
printf("Scenario #%d\n", ++Case);
printf("Frog Distance = %.3f\n\n", f[1][2]);
}
return 0;
}

转载于:https://www.cnblogs.com/wazqWAZQ1/p/4673178.html