HDU1520 Anniversary party(树形DP）

题目链接。

题意：

给定义个多叉树，每个结点上都有一个权值。子结点和父结点不能同时选。求最大的权值和。

分析：

利用左儿子右兄弟原则转化成二叉树，DP。

代码如下：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>

using namespace std;

const int maxn = 6000 + 10;

struct Tree {
    int father, child, brother;
    int Take, Not;

    void init() {
        father = child = brother = Not = 0;
    }
}tree[maxn];

void dfs(int idx) {
    int child = tree[idx].child;
    while(child) {
        dfs(child);
        tree[idx].Take += tree[child].Not;
        tree[idx].Not += max(tree[child].Take, tree[child].Not);
        child = tree[child].brother;
    }
}

int main(){
    int n, x;
    while(scanf("%d", &n) == 1) {
        for(int i=1; i<=n; i++) {
            scanf("%d", &x);
            tree[i].init();
            tree[i].Take = x;
        }

        int L, K;
        while(scanf("%d%d", &L, &K) == 2) {
            if(L == 0 && K == 0) break;
            tree[L].father = K;
            tree[L].brother = tree[K].child;
            tree[K].child = L;
        }

        for(int i=1; i<=n; i++) {
            if(!tree[i].father) {
                dfs(i);
                printf("%d\n", max(tree[i].Take, tree[i].Not));
                break;
            }
        }
    }

    return 0;
}