[ POJ ][ 树形DP ] 2342 Anniversary party

Anniversary party
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 8261   Accepted: 4742

Description

  There is going to be a party to celebrate the 80-th Anniversary of the Ural State University. The University has a hierarchical structure of employees. It means that the supervisor relation forms a tree rooted at the rector V. E. Tretyakov. In order to make the party funny for every one, the rector does not want both an employee and his or her immediate supervisor to be present. The personnel office has evaluated conviviality of each employee, so everyone has some number (rating) attached to him or her. Your task is to make a list of guests with the maximal possible sum of guests' conviviality ratings.

Input

  Employees are numbered from 1 to N. A first line of input contains a number N. 1 <= N <= 6 000. Each of the subsequent N lines contains the conviviality rating of the corresponding employee. Conviviality rating is an integer number in a range from -128 to 127. After that go N – 1 lines that describe a supervisor relation tree. Each line of the tree specification has the form: 
  L K 
It means that the K-th employee is an immediate supervisor of the L-th employee. Input is ended with the line 
  0 0 

Output

  Output should contain the maximal sum of guests' ratings.

Sample Input

7
1
1
1
1
1
1
1
1 3
2 3
6 4
7 4
4 5
3 5
0 0

Sample Output

5

Source

 
首先存下树,记下根节点,然后进行树形dp。
定义:
  dp[i][0] 为 以i为根节点的(子)树 i 不去 时高兴度和的最大值。
  dp[i][1] 为 以i为根节点的(子)树 i 去 时高兴度和的最大值。
  
  先dfs到叶子结点,然后回溯,对于每个节点p, 遍历其son。
  
  dp[p][1] += dp[son[p]][0] ——下属都不能去
  dp[p][0] += max{dp[son[p]][0], dp[son[p]][1]}  ——下属可自由选择去还是不去:max取最优解。
  
  注意在输入 i (i ∈ 1...n) 时直接扫描进dp[i][1]即可。  
//Kvar_ispw17
#include <algorithm>
#include <cstring>
#include <vector>
#include <cstdio>
const int maxn = 6000 + 10;
using namespace std;
vector<int> v[maxn]; int fa[maxn], dp[maxn][2], root, n;
void add(int son, int father) {
	v[father].push_back(son);
	fa[son] = father;
}
void dpfs(int p) {
	vector<int>::iterator iter;
	for(iter = v[p].begin(); iter != v[p].end(); iter++) {
		int s = *iter;
		dpfs(s);
		dp[p][1] += dp[s][0];
		dp[p][0] += max(dp[s][0], dp[s][1]);
	}
}
void print() {
	printf("%d\n", max(dp[root][1], dp[root][0]));
}
int main() {
	while(scanf("%d", &n)) {
		if(!n) break;
		for(int i = 1; i <= n; i++) scanf("%d", &dp[i][1]);
		for(int i = 1; i < n; i++) {
			int sn,f;
			scanf("%d%d", &sn, &f);
			add(sn,f);
		}
		for(int i = 1; i <= n; i++) {
			if(!fa[i]) {
				root = i;
				break;
			}
		}
		dpfs(root);
		print();
		for(int i = 1; i <= n; i++) v[i].clear();
		memset(fa, 0, sizeof fa);
	}
	return 0;
}
posted @ 2017-04-26 13:18  Kvar_ispw17  阅读(189)  评论(0编辑  收藏  举报