poj2342-Anniversary party(树形动规)

#include <iostream>#include <cstring>#include <algorithm>using namespace std;struct poj2342 {/*[问题描述]:大学聚会,要求到场的任何两个人都不是上司/下属关系,每个人都有一个高兴值c[i],可正可负,要求让总的高兴值最大*//*[解题思路]:f[i][0]表示以i为根的子树的最大高兴值且第i个人不去,f[i][1]表示以i为根的子树的最大高兴值且第i个人去则f[i][0]=sum(max(f[k][0],f[k][1])),f[i][1]=c[i]+sum(f[k][0]),k是i的下属*/int n;int root;int c[6005];int f[6005][2];int p[6005];bool vis[6005];void dfs(int node) {vis[node] = true;f[node][1] = c[node];f[node][0] = 0;for (int i = 1; i <= n; i++) {if (!vis[i] && p[i] == node) {dfs(i);f[node][0] += max(f[i][0], f[i][1]);f[node][1] += f[i][0];}}}void work() {cin >> n;for (int i = 1; i <= n; i++) {cin >> c[i];}memset(p, 0, sizeof(p));int x, y;while (cin >> x >> y) {if (x == 0 && y == 0) break;p[x] = y;}root = 1;while (p[root] != 0)root = p[root];//cout << "root:" << root << endl;memset(vis, 0, sizeof(vis));dfs(root);cout << max(f[root][0], f[root][1]) << endl;}};int main(){poj2342 solution;solution.work();return 0;}