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27 lines
1.0 KiB
27 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 = 100010;
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int n; // n个客户
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int sum[N], g[N], h[N]; // 疲劳值前缀和 前i个最大值 后i个最大值
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struct Node {
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int s; // 距离
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int b; // 疲劳
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const bool operator<(const Node &W) {
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return b > W.b;
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}
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} a[N];
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int main() {
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cin >> n;
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for (int i = 1; i <= n; i++) cin >> a[i].s; // 住户i到入口的距离
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for (int i = 1; i <= n; i++) cin >> a[i].b; // 向i住户推销产品会积累的疲劳值
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sort(a + 1, a + 1 + n); // 按疲劳由大到小排序
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for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + a[i].b; // 预算出疲劳值的前缀和
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for (int i = 1; i <= n; i++) g[i] = max(g[i - 1], 2 * a[i].s); // 前i个距离最大值
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for (int i = n; i >= 1; i--) h[i] = max(h[i + 1], 2 * a[i].s + a[i].b); // 后i个距离最大值
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for (int i = 1; i <= n; i++) printf("%d\n", max(sum[i] + g[i], sum[i - 1] + h[i]));
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return 0;
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}
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