Big Event in HDU 多重背包

B - Big Event in HDU

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

多重背包问题，可以将问题转化成01背包。两种思路：将多个相同物品拆分成一个一个价值相同的不同物品；也可以在01背包递推时加一层物品个数的循环。注意这题的坑点！是以一个负整数作为结束，不要想当然以为是-1。因为这个TLE了好久。。以后要认真读题ps：negative integer负整数 positive integer正整数

//第一种写法，耗时1248ms
#include<stdio.h>
#include<string.h>

int f[250005],a[10005];

int max(int x,int y)
{
    return x>y?x:y;
}

int main()
{
    int n,V,sum,c,x,y,i,j;
    while(scanf("%d",&n)&&n>=0){
        memset(f,0,sizeof(f));
        memset(a,0,sizeof(a));
        sum=0;c=0;
        for(i=1;i<=n;i++){
            scanf("%d%d",&x,&y);
            while(y--){
                a[++c]=x;
                sum+=x;
            }
        }
        V=sum/2;
        for(i=1;i<=c;i++){
            for(j=V;j>=a[i];j--){
                f[j]=max(f[j],f[j-a[i]]+a[i]);
            }
        }
        printf("%d %d\n",sum-f[V],f[V]);
    }
    return 0;
}

//第二种写法，耗时811ms
#include<stdio.h>
#include<string.h>

int f[250005],a[1005],b[1005];

int max(int x,int y)
{
    return x>y?x:y;
}

int main()
{
    int n,V,sum,i,j,k;
    while(scanf("%d",&n)&&n>=0){
        memset(f,0,sizeof(f));
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        sum=0;
        for(i=1;i<=n;i++){
            scanf("%d%d",&a[i],&b[i]);
            sum+=a[i]*b[i];
        }
        V=sum/2;
        for(i=1;i<=n;i++){
            for(k=1;k<=b[i];k++){
                for(j=V;j>=0;j--){
                    if(j-a[i]>=0){
                        f[j]=max(f[j],f[j-a[i]]+a[i]);
                    }
                }
            }
        }
        printf("%d %d\n",sum-f[V],f[V]);
    }
    return 0;
}