POJ 2342 Anniversary party

http://poj.org/problem?id=2342

Anniversary party Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3451   Accepted: 1923 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  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 Ural State University Internal Contest October'2000 Students Session

#include"stdio.h"
#include"stdlib.h"
#define   N  6002
#define MAX(a,b) ((a)>(b)? (a):(b))
typedef struct Employee 
{
    int  rank;
    struct Employee *lchild, *nextsibling;
    struct Employee *rightmost; //指向最右孩子节点
    int    id;
}Employee,*Epointer;

Epointer ptrEmployees[N];
int     dp[N][2];

//销毁树，释放空间，防止内存泄露
void freeSpace(int n)
{
   int i;
   for(i=1 ; i<=n; ++i)
   {
       free(ptrEmployees[i]);
   }

}

void clear(int f[],int n)
{
    int i;
    for(i=1 ; i<=n; i++)
    {
        f[i] = 0;
    }
}
void dfs(int parent)
{
    Epointer pparent=ptrEmployees[parent];
    Epointer pchild;
    int child;
    if(pparent->lchild==NULL)
        return ;
    pchild = pparent->lchild;
    while(pchild->nextsibling)
    {
        child = pchild->id;
        dfs(child);
        dp[parent][1] += dp[child][0];
        dp[parent][0] += MAX(dp[child][0],dp[child][1]);
        pchild = pchild->nextsibling;
    }

    child = pchild->id;
    dfs(child);
    dp[parent][1] += dp[child][0];
    dp[parent][0] += MAX(dp[child][0],dp[child][1]);
}




int main()
{
    int n,i,child,parent;
    int f[N]={0};
    while(scanf("%d",&n)!=EOF)
    {
          
          for(i=1 ; i<=n; ++i)
          {
              scanf("%d",&dp[i][1]);
              dp[i][0]=0;
              ptrEmployees[i]=(Epointer)malloc(sizeof(Employee));
              ptrEmployees[i]->lchild = NULL;
              ptrEmployees[i]->nextsibling = NULL;
              ptrEmployees[i]->rank = dp[i][1];
              ptrEmployees[i]->id = i;
          }

          scanf("%d%d",&child,&parent);
          while(child+parent)         //child和parent均为0时输入结束
          {
              if( !ptrEmployees[parent]->lchild )
              {
                  ptrEmployees[parent]->lchild = ptrEmployees[child];
                  ptrEmployees[parent]->rightmost = ptrEmployees[child];
              }
              else
              {
                  ptrEmployees[parent]->rightmost->nextsibling = ptrEmployees[child];
                  ptrEmployees[parent]->rightmost = ptrEmployees[child];
              }
              f[child] = 1;   //child雇员有双亲
              scanf("%d%d",&child,&parent);
              
          }

          //find  president
          for(i=1 ; i<=n; ++i)
          {
              if(f[i]==0)
              {
                  dfs(i);
                  break;
              }
          }

          printf("%d\n",MAX(dp[i][0],dp[i][1]));
          freeSpace(n);
          clear(f,n);
    }
          
    return 0;

}

yinjili

2342

Accepted

628K

141MS

C++

2629B

2013-08-12 00:14:45