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34 lines
1.0 KiB
34 lines
1.0 KiB
2 years ago
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 110;
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int n, x[N], y[N], dist[N], ans[N];
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int main() {
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cin >> n;
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for (int i = 0; i < n; i++) cin >> x[i] >> y[i];
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memset(ans, 0x3f, sizeof ans);
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < n; j++) { // 双重循环枚举每个已知的点
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int X = x[i]; // 假设最终的结果是这个X
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int Y = y[j]; // 假设最终的结果是这个Y
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// 计算每个点到(X,Y)的汉密尔顿距离
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for (int k = 0; k < n; k++)
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dist[k] = abs(x[k] - X) + abs(y[k] - Y);
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// 由小到大排序
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sort(dist, dist + n);
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int cnt = 0;
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for (int k = 0; k < n; k++) {
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cnt += dist[k]; // 累加每一个限定范围的最小距离和
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ans[k] = min(ans[k], cnt);
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}
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}
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}
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// 输出答案
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for (int i = 0; i < n; i++) printf("%d\n", ans[i]);
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return 0;
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}
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