树形DP。用F[k][0]和F[k][1]表示某节点不选和选了之后子树的最大值。那么：
f[i][0]=sigma(max(f[k][0],f[k][1]))
f[i][1]=sigma(f[k][0])+v[i]
解题中用了备忘录。一开始要先找树根。
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <memory.h>
#include <vector>
#define MAX(a, b) a>b?a:b
#define LEN 6005
using namespace std; int F[LEN][2]; // dp state, F[i][0] for i not selected, F[i][1] for i selected
int R[LEN];
bool isRoot[LEN];
int opt(int x, int y, vector<vector<int> > &tree) {
if (F[x][y] != -1) {
return F[x][y];
}
if (y == 0) {
int sum = 0;
for (int i = 0; i < tree[x].size(); i++) {
sum += MAX(opt(tree[x][i], 0, tree), opt(tree[x][i], 1, tree));
}
F[x][y] = sum;
return sum;
}
else if (y == 1) {
int sum = R[x];
for (int i = 0; i < tree[x].size(); i++) {
sum += opt(tree[x][i], 0, tree);
}
F[x][y] = sum;
return sum;
}
else return 0;
} int main()
{
memset(F, 0, sizeof(F));
memset(R, 0, sizeof(R));
memset(isRoot, true, sizeof(isRoot));
int n;
cin >> n;
vector<vector<int> > tree(n+1);
for (int i = 1; i <= n; i++) {
cin >> R[i];
}
int x, y;
for (int i = 1; i <= n-1; i++) {
cin >> x >> y;
tree[y].push_back(x);
isRoot[x] = false;
}
cin >> x >> y; // skip 0, 0 for (int i = 1; i <= n; i++) {
for (int j = 0; j <= 1; j++) {
F[i][j] = -1;
}
}
int root = 0;
for (int i = 1; i <= n; i++) {
if (isRoot[i]) {
root = i;
break;
}
}
int ans = MAX(opt(root, 0, tree), opt(root, 1, tree));
cout << ans << endl;
return 0;
}