POJ2342——Anniversary party【树形dp】

最新推荐文章于 2019-11-24 23:57:32 发布

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#树形dp

本文介绍了一种使用树形动态规划方法解决特定问题的策略，该问题要求在考虑员工及其直接上司关系的情况下，最大化聚会的总快乐值。通过定义状态dp[i][0]和dp[i][1]，分别表示第i个员工参与和不参与时的最大快乐值，文章详细阐述了递归计算过程，并提供了一个完整的C++代码实现。

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Anniversary party POJ - 2342
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

题目大意：
先输入一个数字 $N$ ，代表有 $N$ 个人，其下 $N$ 每行输入一个数字，代表第 $i$ 个人的快乐值，接着每行输入两个数字 $l$ 和 $k$ ，代表 $k$ 是 $l$ 的直接上司，要求在聚会中一个员工和他的直接上司不能同时在场，问能获得的最大快乐值。

解题思路：
这是一道经典的树形dp的题，我们可以用 $d p [i] [0]$ 来代表第 $i$ 个员工来能获得的最大快乐值，用 $d p [i] [1]$ 来代表第 $i$ 个员工不来能获得的最大快乐值，则：
$dp[i][0]=dp[i][0]+∑max(dp[i−1][1],dp[i−1][0])dp[i][0]=dp[i][0]+\sum max(dp[i-1][1],dp[i-1][0])$ 即对其子节点来不来取最大值再求和
$dp[i][1]=dp[i][1]+∑dp[i−1][0]dp[i][1]=dp[i][1]+\sum dp[i-1][0]$ 即对其子节点不来的快乐值求和

代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <set>
#include <utility>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define inf 0x3f3f3f3f
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define lep(i,l,r) for(int i=l;i>=r;i--)
#define ms(arr) memset(arr,0,sizeof(arr))
//priority_queue<int,vector<int> ,greater<int> >q;
const int maxn = (int)1e5 + 5;
const ll mod = 1e9+7;
vector<int> m[6100];
int arr[6100];
int dp[6100][2];
void dfs(int s) {
	if(m[s].size()==0) {
		dp[s][1]=arr[s];
		dp[s][0]=0;
		return;
	}
	dp[s][1]=arr[s];
	dp[s][0]=0;
	for(int i=0;i<m[s].size();i++) {
		dfs(m[s][i]);
		dp[s][1]+=dp[m[s][i]][0];
		dp[s][0]+=max(dp[m[s][i]][1],dp[m[s][i]][0]);
	}
}
bool vis[6100];
int main() 
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    #endif
    //freopen("out.txt", "w", stdout);
    //ios::sync_with_stdio(0),cin.tie(0);
    int n;
    scanf("%d",&n);
    rep(i,1,n) {
    	scanf("%d",&arr[i]);
    }
    int l,k;

    while(scanf("%d %d",&l,&k)&&(l+k)) {
    	m[k].push_back(l);
    	vis[l]=true;
    }
    int i;
    for(i=1;i<=n;i++) {
    	if(vis[i]==false) {
    		dfs(i);
    		break;
    	}
    }
    printf("%d\n",max(dp[i][0],dp[i][1]));
    return 0;
}