2023-9-22 没有上司的舞会

题目链接：没有上司的舞会

#include <cstring>
#include <iostream>
#include <algorithm>using namespace std;const int N = 6010;int n;
int happy[N];
int h[N], e[N], ne[N], idx;
bool has_father[N];// 两个状态，选该节点或不选该节点
int f[N][2];void add(int a, int b)
{e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
}void dfs(int u)
{f[u][1] = happy[u];for(int i = h[u]; i != -1; i = ne[i]){int j = e[i];dfs(j);f[u][0] += max(f[j][0], f[j][1]);f[u][1] += f[j][0];}
}int main()
{scanf("%d", &n);for(int i = 1; i <= n; i ++) scanf("%d", &happy[i]);memset(h, -1, sizeof h);// n 个节点，一共n - 1条边for(int i = 0; i < n - 1; i ++){int a, b;scanf("%d%d", &a, &b);has_father[a] = true;add(b, a);}// 找到根节点int root = 1;while(has_father[root]) root ++;dfs(root);printf("%d\n", max(f[root][0], f[root][1]));return 0;
}