【HDU 1171】【背包DP 或者 母函数】Big Event in HDU【有n样物品，每样物品价值是v，件数是m。尽量把这些物品分成两堆使得两边总价值最接近】

传送门：http://acm.split.hdu.edu.cn/showproblem.php?pid=1171

描述：

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36806    Accepted Submission(s): 12787


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

 

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题意：

有n样物品，每样物品价值是v，件数是m。尽量把这些物品分成两堆使得两边总价值最接近。输出分得的两堆各自的价值。

思路一（多重背包）：

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，可以得出其中一个宿舍所得的最大价值，而另一个宿舍的最大价值也可以相应的得到，而前者必定小于等于后者。

01背包专题可以见Here

代码一：

#include<bits/stdc++.h>
using  namespace std;

int val[5005];
int dp[255555];

int  main(){
int n,sum,cnt;
while(~scanf("%d",&n),n>0){
memset(val, 0, sizeof(val));
memset(dp, 0, sizeof(dp));
sum=0;cnt=0;
int a,b;
for(int i=1; i<=n; i++){
scanf("%d%d",&a,&b);
while(b--){
val[cnt++]=a;//将价值存入数组
sum+=a;
}
}
for(int i=0; i<cnt; i++){
for(int j=sum/2; j>=val[i]; j--){  //01背包
dp[j]=max(dp[j], dp[j-val[i]]+val[i]);
}
}
printf("%d %d\n",sum-dp[sum/2],dp[sum/2]);
}
return 0;
}

思路二（母函数）：

类似于HDU 1085的做法

利用母函数法来解决，因为分成两堆，而两堆中较小的一堆最大为所有物品总价值量的一半，所以母函数的组合数上下就可以设置成总价值量的一半。求出所有的组合后，可以利用贪心的思想来得到答案，因为要求两堆之差尽可能小，所以就可以从总价值量的一半开始向小的方向找，找到最大的价值量，则另一堆的价值量就是总价值量-此堆的价值量。因为组合数可能较大，这里不记录组合种数，而是用一个标记来表示该数能否组合出即可。

代码二：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
using namespace std;

#define N 1000000
int c1
;
int c2
;
int a
; //表示价值
int b
;  //表示个数

int main(){
int n,i,j,k;
while(scanf("%d",&n), n > 0) {
int sum=0;
for(i=0;i<n;i++) {
scanf("%d%d",&a[i],&b[i]);
sum+=a[i]*b[i];
}
int mid=ceil(sum/2);//这个是作为两者的价值分界线
for(i=0;i<=sum/2+10;i++) {
c1[i]=0;c2[i]=0;
}
for(i=0;i<=b[0];i++) {
c1[i*a[0]]=a[0];
}
for(i=1;i<n;i++){
for(j=0;j<=mid;j++) {
for(k=0;k*a[i]+j<=mid&&k<=b[i];k++) {
c2[k*a[i]+j]+=c1[j];
}
}
for(j=0;j<=mid;j++) {
c1[j]=c2[j];
c2[j]=0;
}
}
//这上面是求到了一个关于这一系列设备达到mid之前的各种情况

for(i=mid;i>=0;i--) {
if(c1[i]!=0) break;
}
printf("%d %d\n",sum-i,i);
}
return 0;
}