杭电 HDU ACM 1520 Anniversary party（树形dp 基本）

Anniversary party

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6990    Accepted Submission(s): 3104

Problem 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 T 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

711111111 32 36 47 44 53 50 0

 

Sample Output

5

应该说是首个树形dp题目， 还算挺好理解。

说说最近心情： 总感觉特别迷茫，就是快亚洲区网络预选赛了嘛，出线的几率并不大，但是到九月十三号才开始，选拔赛要打完整个九月，然后也就意味着还得单纯的坚持

acm一个月，可是都大二了 ，我还是想学学别的， 特别是linux，android，java方面，书籍设备都配备好了。 唉~~做完这个题目在刷几个树形dp ，还是重点搞搞搜索。

/*=============================================================================##      Author: liangshu - cbam ##      QQ : 756029571 ##      School : 哈尔滨理工大学 ##      Last modified: 2015-09-01 17:24##     Filename: A.cpp##     Description: #        The people who are crazy enough to think they can change the world, are the ones who do ! =============================================================================*/##include<iostream>#include<sstream>#include<algorithm>#include<cstdio>#include<string.h>#include<cctype>#include<string>#include<cmath>#include<vector>#include<stack>#include<queue>#include<map>#include<set>using namespace std;const int INF = 6032;int fa[INF],vis[INF],dp[INF][2];int n;void dfs(int rt){   vis[rt] = 1;   for(int i = 1; i <= n; i++){       if(!vis[i] && fa[i] == rt){           dfs(i);           dp[rt][1] += dp[i][0];           dp[rt][0] += max(dp[i][0],dp[i][1]);       }   }}int main(){    while(~scanf("%d",&n)){        memset(dp, 0, sizeof(dp));        for(int i = 1; i <= n; i++){            scanf("%d",&dp[i][1]);        }        int root, l, k;        root = 0;        while(scanf("%d%d",&l, &k),l + k){            fa[l] = k;            root = k;        }        memset(vis, 0, sizeof(vis));        dfs(root);        printf("%d\n",max(dp[root][1], dp[root][0]));    }    return 0;}