47 lines
1.4 KiB
C++
47 lines
1.4 KiB
C++
#include<bits/stdc++.h>
|
||
|
||
using namespace std;
|
||
const int INF = 0x3f3f3f3f;
|
||
|
||
const int N = 16; //对于全部的测试点，保证 1<=n<=15，
|
||
int n; //一共多少个奶酪
|
||
double res = INF; //记录最短路径长度,也就是最终的答案
|
||
double dis[N][N]; //dis[i][j]记录第i个点到第j的点的距离.这个是预处理的二维数组，防止重复计算,预处理是搜索优化的重要手段
|
||
|
||
//坐标
|
||
struct Point {
|
||
double x, y;
|
||
} a[N];
|
||
|
||
int per[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11};
|
||
|
||
int main() {
|
||
//老鼠的原始位置
|
||
a[0].x = 0, a[0].y = 0;
|
||
|
||
//读入奶酪的坐标
|
||
cin >> n;
|
||
for (int i = 1; i <= n; i++) cin >> a[i].x >> a[i].y;
|
||
|
||
//预处理
|
||
for (int i = 0; i < n; i++)
|
||
for (int j = i + 1; j <= n; j++) {
|
||
double x1 = a[i].x, y1 = a[i].y, x2 = a[j].x, y2 = a[j].y;
|
||
dis[j][i] = dis[i][j] = sqrt(abs((x1 - x2) * (x1 - x2)) + abs((y1 - y2) * (y1 - y2)));
|
||
}
|
||
|
||
//1~n的全排列，计算每一组组合的距离和，找出最小值,也就能过70%的测试点，11个数据是极限
|
||
double MIN = INF;
|
||
do {
|
||
//计算当前路线下的行走距离
|
||
double s = dis[0][per[0]];
|
||
for (int i = 0; i < n - 1; i++) s += dis[per[i]][per[i + 1]];
|
||
MIN = min(MIN, s);
|
||
} while (next_permutation(per, per + n));
|
||
//输出大吉
|
||
printf("%.2f", MIN);
|
||
return 0;
|
||
}
|
||
|
||
|