Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 26038 | Accepted: 8461 |
Description
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
Output
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
用flody求i到j的每條路徑上的最大步長的最小值!我把三層循環的位置弄反了,。。。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
double dist[300][300];
struct node{
double x, y;
}point[300];
int n;
double get_dist(node a, node b) {
double dx = a.x-b.x;
double dy = a.y-b.y;
return sqrt(dx*dx+dy*dy);
}
int main() {
for (int i = 1; scanf("%d", &n), n; i ++) {
for (int j = 1; j <= n; j ++) {
scanf("%lf%lf", &point[j].x, &point[j].y);
}
for (int j = 1; j <= n; j ++) {
for (int k = j; k <= n; k ++) {
dist[j][k] = dist[k][j] = get_dist(point[j], point[k]);
}
}
for (int j = 1; j <= n; j ++) {
for (int k = 1; k <= n; k ++) {
for (int t = 1; t <= n; t ++) {
if (dist[k][t] > max(dist[k][j], dist[j][t])) {
dist[k][t] = max(dist[k][j], dist[j][t]);
}
}
}
}
printf("Scenario #%d\n", i);
printf("Frog Distance = %.3f\n\n", dist[1][2]);
}
return 0;
}