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描述：
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 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.
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.
代码：
多重背包转化为01背包问题，可以认为在一个容量为所有物品价值累加和的一半的背包中，尽量达到最大值。
第一版：
#include<stdio.h> #include<string.h> #include<iostream> #include<stdlib.h> #include <math.h> #define N 105 #define M 1000 using namespace std; int main(){ int n,v[N],dp[N][M],a,b,volume,sum; while( scanf("%d",&n)!=EOF && n>0 ){ volume=0;sum=1; for( int i=0;i<n;i++ ){ scanf("%d%d",&a,&b); volume+=(a*b); while( b-- )//将单个物品依次加入 v[sum++]=a; } sum--; //需要初始化dp中的全部值为0，下一次dp比较的值，上一次可能没有赋值，比较时将与负值（未初始化）进行比较，得到不正确的值 //for( int i=0;i<volume;i++ ) // dp[0][i]=0; memset(dp,0,sizeof(dp)); for( int i=1;i<=sum;i++ ){ for( int j=v[i];j<=volume/2;j++ ){ dp[i][j]=max(dp[i-1][j],dp[i-1][j-v[i]]+v[i]);//转移方程 } } printf("%d %d\n",max(dp[sum][volume/2],volume-dp[sum][volume/2]),min(dp[sum][volume/2],volume-dp[sum][volume/2])); } system("pause"); return 0; }
测试用例：
2
10 1
20 1
3
10 1
20 2
30 1
-1
测试用例2答案是对的，用例1答案不对。因为对于用例1,20大于总价值一半（15），故dp值均为0，最后结果为0。为此，不妨使用优化版的dp，即使用一维数组dp，这样就算出现刚才的情况，较早的dp值即为答案。
#include<stdio.h> #include<string.h> #include<iostream> #include<stdlib.h> #include <math.h> #define N 105 #define M 100000 using namespace std; int main(){ int n,v[N],dp[M],a,b,volume,sum; while( scanf("%d",&n)!=EOF && n>0 ){ volume=0;sum=1; for( int i=0;i<n;i++ ){ scanf("%d%d",&a,&b); volume+=(a*b); while( b-- )//将单个物品依次加入 v[sum++]=a; } sum--; //需要初始化dp中的全部值为0，下一次dp比较的值，上一次可能没有赋值，比较时将与负值（未初始化）进行比较，得到不正确的值 memset(dp,0,sizeof(dp)); for( int i=1;i<=sum;i++ ){ for( int j=volume/2;j>=v[i];j-- ){ dp[j]=max(dp[j],dp[j-v[i]]+v[i]);//转移方程 } } printf("%d %d\n",max(dp[volume/2],volume-dp[volume/2]),min(dp[volume/2],volume-dp[volume/2])); } system("pause"); return 0; }
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原文地址：http://www.cnblogs.com/lucio-yr/p/4722518.html
