HDU 1520(树形DP)
Anniversary party
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9137 Accepted Submission(s): 3917
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
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
树形DP和线性DP差不多. dfs一下。
#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <vector> using namespace std; int n; int happy[6005]; vector<int> son[6005]; int dp[6005][3]; int vis[6005]; void dfs(int now) { dp[now][1] = happy[now]; dp[now][0] = 0; for(int i=0;i<son[now].size();i++) { int nex = son[now][i]; dfs(nex); dp[now][1] += dp[nex][0]; dp[now][0] += max(dp[nex][0],dp[nex][1]); } } int main() { while(scanf("%d",&n)!=EOF&&n) { memset(dp,0,sizeof(dp)); memset(vis,0,sizeof(vis)); for(int i=1;i<=n;i++) { scanf("%d",&happy[i]); } int l,k; while(1) { scanf("%d %d",&l,&k); if(l==0&&k==0) break; son[k].push_back(l); vis[l] = 1; } int father = 0; for(int i=1;i<=n;i++) { if(!vis[i]) { father = i; break; } } dfs(father); printf("%d\n",max(dp[father][0],dp[father][1])); for(int i=1;i<=n;i++) son[i].clear(); } return 0; }
分类:
ACM-OJ-HDU
, ACM-动态规划-树形DP
【推荐】凌霞软件回馈社区,博客园 & 1Panel & Halo 联合会员上线
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步
· 智能桌面机器人:用.NET IoT库控制舵机并多方法播放表情
· Linux glibc自带哈希表的用例及性能测试
· 深入理解 Mybatis 分库分表执行原理
· 如何打造一个高并发系统?
· .NET Core GC压缩(compact_phase)底层原理浅谈
· 开发者新选择:用DeepSeek实现Cursor级智能编程的免费方案
· Tinyfox 发生重大改版
· 独立开发经验谈:如何通过 Docker 让潜在客户快速体验你的系统
· 小米CR6606,CR6608,CR6609 启用SSH和刷入OpenWRT 23.05.5
· 近期最值得关注的AI技术报告与Agent综述!