HDU1171-Big Event In HDU(母函数、动态规划）

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16416    Accepted Submission(s): 5792

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
   
   
   
   

 

题目大意：给定一些物品，每种物品都有固定价值和数量，现在要把这些物品尽可能拼接分成两份（单个物品不能切割），请输出分割后各份的值（如果不一般大，则先输出大的）。

 

分析：

思路一：我们可以利用母函数，建立该模型的母函数，算出x^(sum/2)的系数即可！代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int c1[125100],c2[125100],f[55],v[55];

int main() {
    int n,i,j,k,aim,sum;
    while(scanf("%d",&n),n>=0) {
        for(i=1,sum=0;i<=n;i++) {
            scanf("%d %d",&v[i],&f[i]);
            sum+=f[i]*v[i];
        }
        aim=sum/2;
        memset(c1,0,sizeof(c1));
        memset(c2,0,sizeof(c2));
        
        for(j=0,k=0;k<=f[1];k++,j+=v[1])//第一个括号里面只能取到这么多 
            c1[j]=1;
           
        for(i=2;i<=n;i++) {
            for(j=0;j<=aim;j++) {
                if(c1[j]==0) continue;
                
                for(k=0;k*v[i]+j<=aim&&k<=f[i];k++)
                    c2[k*v[i]+j]=c1[j];
            }
            for(j=0;j<=aim;j++) {
                c1[j]=c2[j];
                c2[j]=0;
            }
        }
        for(i=aim;i>=0;i--)
            if(c1[i]!=0) break;
        printf("%d %d\n",sum-i,i);
    }
    return 0;
}

思路二：利用动态规划之多重背包问题，我们可以看成是背包问题，背包容量是总价值的一般，目标是挑出一些物品，尽可能装满背包。

方法一：转化成0-1背包来做。代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int dp[126000],v[60],w[110];
int main(){
    int n,i,j,k,sum;
    while(scanf("%d",&n),n>=0) {
        for(i=0,sum=0;i<n;i++) {
            scanf("%d %d",&v[i],&w[i]);
            sum+=v[i]*w[i];
        }
        memset(dp,0,sizeof(dp));
        for(i=0;i<n;i++)
            for(j=1;j<=w[i];j++)
                for(k=sum/2;k>=v[i];k-=v[i])
                    if(dp[k-v[i]]+v[i]>dp[k])
                        dp[k]=dp[k-v[i]]+v[i];
        printf("%d %d\n",sum-dp[sum/2],dp[sum/2]);
    }
    return 0;
}

方法二：将每种物品打包成价值为1、2、4、。。等，加快挑选过程！代码：

#include<stdio.h>
#include<string.h>
#define N 55
#define M 1250010
int dp[M],f[N+1],v[N+1];
int main(){
    int i,j,k,flag,sum,s,n;
    while(scanf("%d",&n),n>=0) {
        for(i=1,sum=0;i<=n;i++){
            scanf("%d %d",&v[i],&f[i]);
            sum+=f[i]*v[i];
        }
        memset(dp,0,sizeof(dp[0])*(sum/2+10));
        for(i=1;i<=n;i++){
            for(j=sum/2;j>=v[i];j--) {
                for(k=1,s=0;k<=f[i]&&j>=k*v[i];k*=2) { 
                    dp[j]=dp[j-k*v[i]]+k*v[i]>dp[j]?dp[j-k*v[i]]+k*v[i]:dp[j];
                }
                k=f[i]-s;
                if(j>=k*v[i]) dp[j]=dp[j-k*v[i]]+k*v[i]>dp[j]?dp[j-k*v[i]]+k*v[i]:dp[j];
            }
        }
        printf("%d %d\n",sum-dp[sum/2],dp[sum/2]);
    }
    return 0;
}