HDU1171 Big Event in HDU(01背包的转化，多重背包)

题目：

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 37483    Accepted Submission(s): 13007


Problem 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

2
10 1
20 1
3
10 1
20 2
30 1
-1

 

Sample Output

20 10
40 40

 

Author

lcy

 

代码1（转化成01背包）：

因为题目要求要尽量平均分配，所以我们可以先将总价值sum求出，然后得出其分配的平均值为sum/2,要注意这个答案可能为小数，但是又因为sum是整数，所以最后得出的sum/2是要小于等于实际的值。将这个结果进行01,背包，可以得出其中一个学院所得的最大价值，而另一个学院的最大价值也可以相应的得到，而前者必定小于等于后者。和邮票分你一半，zb的生日也比较相似

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
int a[5005];
int dp[250005];
int main()
{
int n;
while(1)
{
//sum的一半一定是
scanf("%d",&n);
if(n<0)return 0;//为负数的时候结束输入
mem(a,0);
mem(dp,0);//初始化
int sum=0,l=0;
int value,num;//价值和数量
for(int i=0; i<n; i++)
{
scanf("%d %d",&value,&num);
for(int j=0;j<num;j++)
{
a[l++]=value;//把价值存入数组，转化成01背包问题
sum+=value;//将sum求和
}
}
for(int i=0; i<l; i++)//以一半的sum进行01背包
for(int j=sum/2; j>=a[i]; j--)
dp[j]=max(dp[j],dp[j-a[i]]+a[i]);
printf("%d %d\n",sum-dp[sum/2],dp[sum/2]);
}
return 0;
}

方法2（多重背包）：

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
int value[5005];
int num[5005];
int dp[250005];
int main()
{
int n;
while(1)
{
//sum的一半一定是
scanf("%d",&n);
if(n<0)return 0;//为负数的时候结束输入
mem(value,0);
mem(num,0);
mem(dp,0);//初始化
int sum=0;

for(int i=0; i<n; i++)
{
scanf("%d %d",&value[i],&num[i]);
sum+=value[i]*num[i];//算出总价值
}
for(int i=0; i<n; i++)//进行多重背包
for(int j=1; j<=num[i]; j++)
for(int k=sum/2; k>=value[i]; k--)
dp[k]=max(dp[k],dp[k-value[i]]+value[i]);
printf("%d %d\n",sum-dp[sum/2],dp[sum/2]);
}
return 0;
}


ps:多重背包的二进制分解暂时还不会，先放着~