HDU 1520 Anniversary party

Anniversary party

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9128    Accepted Submission(s): 3915

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

 

AC的第一道树形dp，一开始由于使用了四次memset函数导致程序运行内存超限，改为一次后AC了，（若不用memset，初始化可能会浪费时间！）但空间复杂度还是很悬啊。。。此后应该谨慎使用memset啊。。。

状态方程:(dp[x][0]表示x不参加的最优解，dp[x][1]表示x参加的最优解,y为其下属）

               dp[x][0]+=max(dp[y][0],dp[y][1]);

               dp[x][1]+=dp[y][0];(x去了下属就不可以去);

p.s：个人认为员工与其下属关系的存储很巧妙！（也可能我太菜。。。）

Problem : 1520 ( Anniversary party )     Judge Status : Accepted
RunId : 17972490    Language : C++    Author : 15030120010
Code Render Status : Rendered By HDOJ C++ Code Render Version 0.01 Beta

#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;int dp[6001][2];int yg[6005][6005];int h[6001];bool ss[6001]; void dfs(int x){ int i,y; dp[x][0]=0; dp[x][1]=h[x]; for(i=1;i<=yg[x][0]&&yg[x][i];++i){ y=yg[x][i]; dfs(y); dp[x][0]+=max(dp[y][0],dp[y][1]); dp[x][1]+=dp[y][0]; }}int main(){ int n,i,j,x,y; while(scanf("%d",&n)!=EOF){ memset(dp,0,sizeof(dp)); for(i=1;i<=n;++i){ scanf("%d",&h[i]); yg[i][0]=0; ss[i]=0; } while(scanf("%d %d",&x,&y),x||y){ yg[y][++yg[y][0]]=x;//yg[y][0]表示y的下属的个数 ++ss[x];//ss[y]表示y的上级个数 } i=0; while(ss[++i]); dfs(i); printf("%d\n",max(dp[i][1],dp[i][0])); } return 0;}

#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;int dp[6001][2];int yg[6005][6005];int h[6001];bool ss[6001]; void dfs(int x){    int i,y;    dp[x][0]=0;    dp[x][1]=h[x];    for(i=1;i<=yg[x][0]&&yg[x][i];++i){        y=yg[x][i];        dfs(y);        dp[x][0]+=max(dp[y][0],dp[y][1]);        dp[x][1]+=dp[y][0];    }}int main(){    int n,i,j,x,y;    while(scanf("%d",&n)!=EOF){        memset(dp,0,sizeof(dp));        for(i=1;i<=n;++i){            scanf("%d",&h[i]);            yg[i][0]=0;            ss[i]=0;        }        while(scanf("%d %d",&x,&y),x||y){            yg[y][++yg[y][0]]=x;//yg[y][0]表示y的下属的个数            ++ss[x];//ss[y]表示y是否有上级        }        i=0;        while(ss[++i]);        dfs(i);        printf("%d\n",max(dp[i][1],dp[i][0]));    }    return 0;}