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#include<bits/stdc++.h>
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using namespace std;
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const int INF = 0x3f3f3f3f;
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const int N = 16; //对于全部的测试点,保证 1<=n<=15,
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int n; //一共多少个奶酪
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double res = INF; //记录最短路径长度,也就是最终的答案
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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
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//坐标
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struct Point {
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double x, y;
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} a[N];
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int per[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11};
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int main() {
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//老鼠的原始位置
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a[0].x = 0, a[0].y = 0;
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//读入奶酪的坐标
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cin >> n;
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for (int i = 1; i <= n; i++) cin >> a[i].x >> a[i].y;
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//预处理
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for (int i = 0; i < n; i++)
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for (int j = i + 1; j <= n; j++) {
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double x1 = a[i].x, y1 = a[i].y, x2 = a[j].x, y2 = a[j].y;
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dis[j][i] = dis[i][j] = sqrt(abs((x1 - x2) * (x1 - x2)) + abs((y1 - y2) * (y1 - y2)));
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}
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//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
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double MIN = INF;
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do {
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//计算当前路线下的行走距离
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double s = dis[0][per[0]];
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for (int i = 0; i < n - 1; i++) s += dis[per[i]][per[i + 1]];
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MIN = min(MIN, s);
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} while (next_permutation(per, per + n));
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//输出大吉
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printf("%.2f", MIN);
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return 0;
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}
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