【树形dp+并查集】Anniversary party POJ

Think:
1题意：举行聚会，参会人选不希望自己的直属上司参加，关系形成一棵树，参会人选每个人有自己的开心度，求最大开心度
２思路：树形dp + 并查集（寻找根节点）

建议参考博客

以下为Accepted代码

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 6004;int n, dp[N][2], far[N], vis[N];void DFS(int u);int main(){    int i, s, f, root;    while(~scanf("%d", &n)){        memset(dp, 0, sizeof(dp));        memset(vis, 0, sizeof(vis));        for(i = 1; i <= n; i++)/*并查集初始化*/            far[i] = i;        for(i = 1; i <= n; i++)            scanf("%d", &dp[i][1]);        while(scanf("%d %d", &s, &f) && (s || f)){            far[s] = f;        }        root = 1;        while(far[root] != root)/*寻找根节点*/            root = far[root];        DFS(root);        printf("%d\n", max(dp[root][1], dp[root][0]));    }    return 0;}void DFS(int u){    vis[u] = 1;    for(int i = 1; i <= n; i++){        if((!vis[i]) && far[i] == u){            DFS(i);            dp[u][1] += dp[i][0];/*上属去下属不去*/            dp[u][0] += max(dp[i][1], dp[i][0]);/*上属不去前提下下属去或不去取最大值*/        }    }}