POJ2342:Anniversary party(树形DP)
来源:互联网 发布:js图片热点区域 编辑:程序博客网 时间:2024/04/29 20:01
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
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
711111111 32 36 47 44 53 50 0
Sample Output
5
题意:有n个人,接下来n行是n个人的价值,再接下来n行给出l,k说的是l的上司是k,这里注意l与k是不能同时出现的
思路:用dp数据来记录价值,开数组用下标记录去或者不去、
则状态转移方程为:
DP[i][1] += DP[j][0],
DP[i][0] += max{DP[j][0],DP[j][1]};其中j为i的孩子节点。
这样,从根节点r进行dfs,最后结果为max{DP[r][0],DP[r][1]}。
#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int father[6005],vis[6005],dp[6005][2],t;void dfs(int node){ int i,j; vis[node] = 1; for(i = 1;i<=t;i++) { if(!vis[i] && father[i] == node) { dfs(i); dp[node][1]+=dp[i][0];//node去,则i必不能去 dp[node][0]+=max(dp[i][0],dp[i][1]);//node不去,取i去或不去的最大值 } }}int main(){ int i,j,l,k,root; while(~scanf("%d",&t)) { for(i = 1;i<=t;i++) scanf("%d",&dp[i][1]); root = 0; while(scanf("%d%d",&l,&k),l+k>0) { father[l] = k;//记录上司 root = k; } memset(vis,0,sizeof(vis)); dfs(root); printf("%d\n",max(dp[root][1],dp[root][0])); } return 0;}
- POJ2342:Anniversary party(树形DP)
- poj2342 Anniversary party【树形dp】
- POJ2342 Anniversary party(树形DP)
- poj2342.Anniversary party(树形dp)
- POJ2342 Anniversary party 【树形dp】
- poj2342 Anniversary party (树形DP)
- POJ2342 Anniversary party(树形dp)
- poj2342|hdu 1520 Anniversary party 树形dp
- hdu 1520 && poj2342 anniversary party树形DP
- Hdoj 1520&Poj2342 Anniversary party 【树形DP】
- POJ2342 Anniversary party 树形dp入门题
- poj2342 Anniversary party(树形dp)
- hdu1520 Anniversary party(poj2342,树形dp)
- URAL 1039 / poj2342-Anniversary Party-树形DP
- [树形dp][入门]hdu1520 & poj2342 Anniversary party
- POJ2342 ->Anniversary party(树形DP入门题)
- poj2342 Anniversary party--树形dp入门
- POJ2342:Anniversary party(树形dp)
- C#Random
- android Service(二) activity启动Service方式二:bindService()
- UVC 1.0 和 UVC 1.1的差别
- Android源码中增加模块
- asp实现语音上传
- POJ2342:Anniversary party(树形DP)
- C#基础——类型转换
- 使用FTP出现的3种常见错误
- 利用容器实现字符串的权值
- mvc与三层架构不得不说的秘密
- vlc的应用之九:用vlc串流摄像头
- 用VBS代码实现简单的场景恢复实例
- Square_hdu_1518(深搜)
- vector