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57 lines
1.3 KiB
57 lines
1.3 KiB
#include <bits/stdc++.h>
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using namespace std;
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const int N = 100010;
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vector<int> g[N], f[N]; // 邻接表,一个正图，一个反图
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int c[N]; // 每个树洞中松鼠的数量
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/*
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测试用例
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4
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5
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3
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6
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1
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1 2
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1 3
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2 4
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2
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*/
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int dfs1(int u, int k) {
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if (k == 0) return 0;
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int sum = c[u];
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for (int i = 0; i < g[u].size(); i++) sum += dfs1(g[u][i], k - 1);
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return sum;
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}
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int dfs2(int u, int k) {
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if (k == 0) return 0;
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int sum = c[u];
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for (int i = 0; i < f[u].size(); i++) sum += dfs2(f[u][i], k - 1);
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return sum;
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}
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int main() {
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int n; // 树洞的数量
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cin >> n;
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for (int i = 1; i <= n; i++) cin >> c[i]; // 每个树洞中松鼠的数量
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for (int i = 1; i < n; i++) {
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int a, b; // 表示两个树洞相连接
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cin >> a >> b;
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g[a].push_back(b); // 用邻接表保存一下某个树洞与哪两个其它树洞相连接
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f[b].push_back(a); // 建反图
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}
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int k;
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cin >> k;
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for (int i = 1; i <= n; i++) {
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// 计算i号树洞与它在k个距离内的所有树洞的松鼠个数和
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// 1、向下k个距离 k<=20
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int sum = dfs1(i, k);
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// 2、向上k个距离 k<=20
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sum += dfs2(i, k);
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cout << sum - c[i] << endl;
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}
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return 0;
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} |