zoukankan      html  css  js  c++  java
  • HDOJ 题目1520 Anniversary party(树形dp)

    Anniversary party

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 6271    Accepted Submission(s): 2820


    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
    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
     

    Recommend
    linle   |   We have carefully selected several similar problems for you:  2196 1494 2242 3001 1074 
    题目大意:有n个人去參加party,每一个人都有自己的价值。后边跟很多行a b(0,0结束),表示b是a的上司,不能使他和他的上司同一时候參加,能够获得最大的快乐值,
    dp[i][0]表示他不參加的最大快乐值,dp[i][1]表示他參加最大的快乐值
     ac代码
    #include<stdio.h>
    #include<string.h>
    #define max(a,b) (a>b?

    a:b) struct s { int u,v,next; }edge[3000100]; int a[6060],dp[6060][2],dig[6060],head[6060],vis[6060],cnt; void add(int u,int v) { edge[cnt].u=u; edge[cnt].v=v; edge[cnt].next=head[u]; head[u]=cnt++; } void dfs(int u) { vis[u]=1; dp[u][1]=a[u]; int i; for(i=head[u];i!=-1;i=edge[i].next) { int v=edge[i].v; if(!vis[v]) { dfs(v); dp[u][0]+=max(dp[v][0],dp[v][1]); dp[u][1]+=dp[v][0]; } } } int main() { int n; while(scanf("%d",&n)!=EOF) { int i; for(i=1;i<=n;i++) { scanf("%d",&a[i]); } int u,v; cnt=0; memset(head,-1,sizeof(head)); memset(dp,0,sizeof(dp)); memset(dig,0,sizeof(dig)); memset(vis,0,sizeof(vis)); while(scanf("%d%d",&v,&u),u||v) { add(u,v); dig[v]++; } int s; for(i=1;i<=n;i++) { if(!dig[i]) { s=i; } } dfs(s); printf("%d ",max(dp[s][0],dp[s][1])); } }




  • 相关阅读:
    ci中使用mail类
    简单php发送邮件
    Firefox 中出现的 “Network Protocol Error”怎么办
    让linux启动更快的方法
    小米盒子3代码公布了,你要刷机吗?
    毕业季,我的Linux求职之路
    菜鸟学习计划浅谈之Linux系统
    细述:nginx http内核模块提供的变量和解释
    Linux使用百度云
    网工的Linux系统学习历程
  • 原文地址:https://www.cnblogs.com/cxchanpin/p/7056428.html
Copyright © 2011-2022 走看看