ACM: 动态规划题+剪枝 toj 3904

                 
Points cakes

描述

 
A mother has n different
sizes of cake,she wants to give to his two
children by a fair way(which may not completely average, but
requires minimum), 

you can help she solve this problem?

输入

 
there are many case of
testdata,for each testdata,

first line: an integer
n(3<=n<=1000), means how many cakes
does the mother has.

second line: n integers, for each
a[i](1<=a[i]<=50), a[i] means the
size of ith cake.

输出

 
for each teatdata, output
two integers x, y(x<=y)

x, y means how many cakes do the two children have.

样例输入
3

1 2 3

样例输出
3
3

题意: 分配蛋糕, 有n块不同大小的蛋糕, 要求分配给两个人, 如果不可以实现分配均匀(重量相同分配).

      尽量达到大小相近.

解题思路:

       
1. 动态规划: 决策1: 使用当前块, w+a[i] <= sum_half;
==> DP(w+a[i],i+1);

                     决策2:
不使用当前块, 直接跳过. DP(w,i+1);

       
2. 纠结一天问题: (1).递归出口是 i == n+1; (一开始忽略了最后一块i==n是错误的)

                         (2).剪枝:
if(i == n+1 || maxt == sum_half) 递归出口增加.

                                  
已经满足条件的不需要再将递归树伸展了. (TLE解决)

代码:

#include <cstdio>

#include <iostream>

#include <cstring>

#include <cstdlib>

using namespace std;

#define MAX 1005

int n;

int a[MAX];

int maxt, sum;

int sum_half;

void DP(int w,int reLen)

{

 if(reLen == n+1 || maxt == sum_half)

 {

  if(maxt < w
&& w <= sum_half)
maxt = w;

  return ;

 }

 if(w+a[reLen] <= sum_half)
DP(w+a[reLen],reLen+1);

 DP(w,reLen+1);

}

int main()

{

 int i, j;

// freopen("input.txt","r",stdin);

 while(scanf("%d",&n) !=
EOF)

 {

  sum = 0;

  for(i = 1; i <=
n; ++i)

  {

   scanf("%d",&a[i]);

   sum +=
a[i];

  }

  maxt = 0;

  sum_half = (sum%2==0 ? sum/2 :
sum/2+1);

  DP(0,1);

  int big = (maxt
> sum-maxt ? maxt : sum-maxt);

  printf("%d
%d\n",sum-big,big);

 }

 return 0;

}