杭電1171 Big Event in HDU（母函數+多重背包解法）

Big Event in HDU

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

Total Submission(s): 23728    Accepted Submission(s): 8363

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
          
/*
母函數解法：将能籌成的标記成1，找最接近sum/2的就可以
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using std::max;
const int MAX=52;
bool c1[MAX*MAX*100];//c2[MAX*MAX*100];
int main(){
	int N;
	int val[MAX],Count[MAX];
	while(scanf("%d",&N),N>0){
		int sum=0;
		memset(val,0,sizeof(val));
		memset(Count,0,sizeof(Count));
		memset(c1,0,sizeof(c1));
		for(int i=1;i<=N;i++){
		scanf("%d%d",&val[i],&Count[i]);
		sum+=val[i]*Count[i];
		}
		c1[0]=1;
		int Limit=sum>>1;
		for(int i=1;i<=N;i++){
			for(int j=0;j<=Limit;j++){
				for(int k=0;j+k<=Limit&&k<=val[i]*Count[i];k+=val[i])
				c1[k+j]=(c1[k+j]==1?c1[k+j]:c1[j]);//注意此處不能直接等于，防止出現 50 30 20 ，v[20]=0;的情況，是可以籌成50的
			}
			/*
			for(int j=0;j<=Limit;j++){
				c1=c2[j];
				c2[j]=0;
			}*/
		}
		int k=Limit;
		while(1){
			if(c1[k]){
			printf("%d %d\n",sum-k,k);
			break;
			}
			k--;
		}
	}
return 0;
} 
           
    
/*
多重背包解法：
将體積和價值都看做 價值，上限也是 sum/2，找最大值，用sum-d[sum/2]便是較大的那一個
*/
#include<stdio.h>
#include<string.h>
#include<algorithm>
using std::max;
const int MAX=51;
int sum,Count,N;
int dp[MAX*MAX*100];
int v[MAX],w[MAX],c[MAX];
int V[MAX],W[MAX];
void Div(){//多重背包分解 
	Count=0;
	for(int i=0;i<N;i++){
		for(int j=1;j<=c[i];j<<=1){
			W[Count++]=w[i]*j;
			c[i]-=j;
		}
		if(c[i]>0){
			W[Count++]=w[i]*c[i];
		}
	}
}
int main(){
	while(scanf("%d",&N),N>0){
		sum=0;
		memset(w,0,sizeof(w));
		memset(c,0,sizeof(c));
		memset(W,0,sizeof(W));
		memset(dp,0,sizeof(dp));
		for(int i=0;i<N;i++){
		scanf("%d%d",&w[i],&c[i]);
		sum+=w[i]*c[i];
		}
		int upLimit=sum>>1;
			Div();
		for(int i=0;i<Count;i++){
			for(int j=upLimit;j>=W[i];j--){
				dp[j]=max(dp[j],dp[j-W[i]]+W[i]);
			}
		}
		printf("%d %d\n",sum-dp[upLimit],dp[upLimit]);
	}
return 0;
}