hdu 1117 Big Event in HDU（多重背包）

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概述

这题目和完全背包问题很类似。基本的方程只需将完全背包问题的方程略微一改即可，因为对于第i种物品有n[i]+1种策略：取0件，取1件……取n[i]件。令f[i][v]表示前i种物品恰放入一个容量为v的背包的最大权值，则有状态转移方程：

f[i][v]=max{f[i-1][v-k*c[i]]+k*w[i]|0<=k<=n[i]}

Description

Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been split into Computer College and Software College in 2002.
The splitting is absolutely a big event in HDU! At the same time, it is a trouble thing too. All facilities must go halves. First, all facilities are assessed, and two facilities are thought to be same if they have the same value. It is assumed that there is N (0<N<1000) kinds of facilities (different value, different kinds).

Input

Input contains multiple test cases. Each test case starts with a number N (0 < N <= 50 -- the total number of different facilities). The next N lines contain an integer V (0<V<=50 --value of facility) and an integer M (0<M<=100 --corresponding number of the facilities) each. You can assume that all V are different.
A test case starting with a negative integer terminates input and this test case is not to be processed.

Output

For each case, print one line containing two integers A and B which denote the value of Computer College and Software College will get respectively. A and B should be as equal as possible. At the same time, you should guarantee that A is not less than B.

Sample Input

Sample Output

20 10
40 40

这是多重背包，背包的上线时total/2;

代码：

#include<iostream>
using namespace std;
int dp[250025];
int v[100],num[100];
int max( int a,int b ) 
{ 
    return a > b ? a : b; 
} 
int main()
{
    int i,j,n,sum,cnt,k;
    while(scanf("%d",&n),n>0)
    {
      cnt=0;
      for(i=0;i<n;i++){ scanf("%d%d",&v[i],&num[i]);cnt+=v[i]*num[i];}
      memset(dp,0,sizeof(dp));
      sum=cnt/2;
      for(i=0;i<n;i++)
        for(j=0;j<num[i];j++)
           for(k=sum;k>=v[i];k-=v[i])                 
             dp[k]=max(dp[k],dp[k-v[i]]+v[i]);
      if(cnt-dp[sum]>dp[sum])
       printf("%d %dn",cnt-dp[sum],dp[sum]);
       else printf("%d %dn",dp[sum],cnt-dp[sum]);
    }
    return 0;
}