POJ 2342 Anniversary party (树形dp 入门题)

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
int const MAX = 6005;
int dp[MAX][2], val[MAX];
bool vis[MAX], flag[MAX];
vector <int> vt[MAX];

void DFS(int fa)
{
    vis[fa] = true;
    dp[fa][1] = val[fa];
    int sz = vt[fa].size();
    for(int i = 0; i < sz; i++)
    {
        int son = vt[fa][i];
        if(!vis[son])
        {
            DFS(son);
            dp[fa][1] += dp[son][0];
            dp[fa][0] += max(dp[son][1], dp[son][0]);
        }
    }
    return;
}

int main()
{
    int n;
    while(scanf("%d", &n) && n)
    {
        for(int i = 1; i <= n; i++)
            vt[i].clear();
        memset(flag, false, sizeof(flag));
        memset(vis, false, sizeof(vis));
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; i++)
            scanf("%d", &val[i]);
        for(int i = 0; i < n - 1; i++)
        {
            int a, b;
            scanf("%d %d", &a, &b);
            vt[b].push_back(a);
            flag[a] = true;
        }
        int root;
        for(int i = 1; i <= n; i++) //n个点n－1条边，必然存在“入度”为0的点即树根
        {
            if(!flag[i])
            {
                root = i;
                break;
            }
        }
        DFS(root);
        printf("%d\n", max(dp[root][1], dp[root][0]));
    }
}