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hdu1171Big Event in HDU (多重背包)

2018年02月22日 ⁄ 综合 ⁄ 共 1928字 ⁄ 字号 评论关闭
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
题目意思:有n台设备,每台设备有独一无二的价值和设备的数量。要求总价值尽可能的平分给Computer College 和 Software College。输出时要求第一个数大于等于第二个数。
解题:用多重背包做出。注意在杭电上判断是否结束输入时,条件为n<0.
#include<stdio.h>
#define N 1000100
int dp[N],vect;
int max(int a,int b){return a>b?a:b;}
void zeroonepack(int use,int w)
{
    for(int v=vect;v>=use;v--)
    dp[v]=max(dp[v],dp[v-use]+w);
}
void complexepack(int use,int w)
{
    for(int v=use;v<=vect;v++)
    dp[v]=max(dp[v],dp[v-use]+w);
}
void mulitpack(int use,int w,int m)
{
    if(m*use>=vect)
        complexepack(use,w);
    else
    {
        int k=1;
        while(k<m)
        {
            zeroonepack(k*use,k*w);
            m-=k; k*=2;
        }
        zeroonepack(m*use,m*w);
    }
}
struct nn
{
    int w,m;
}node[N];
int main()
{
    int n,A,B,sum;
    while(scanf("%d",&n)>0&&n>0)
    {
        vect=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&node[i].w,&node[i].m);
            vect+=node[i].w*node[i].m;
        }
        sum=vect;
        vect/=2;
        for(int i=0;i<=vect;i++)
            dp[i]=0;
        for(int i=1;i<=n;i++)
        mulitpack(node[i].w,node[i].w,node[i].m);
        A=max(dp[vect],sum-dp[vect]);
        B=sum-A;
        printf("%d %d\n",A,B);
    }
}

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