HDU 1171 Big Event in HDU（背包）

最新推荐文章于 2025-09-10 22:21:01 发布

转载最新推荐文章于 2025-09-10 22:21:01 发布 · 58 阅读

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#java

本文介绍了一道关于将设施平均分配给两个学院的问题，并通过多重背包优化的方式进行解答。使用C++实现，对比了母函数和背包算法的区别，提供了一个高效的解决方案。

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16224 Accepted Submission(s): 5725

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

题目意思就是n种东西，知道价值和数量。问尽量分成两份，两份对应的价值，先输出大的，再输出小的。

以前用母函数做的，今天做DP复习了下，貌似直接用多重背包更加简单。想想以前的母函数的题目，貌似很多时候都可以换成背包做的

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <queue>
#include <string>
#include <set>
#include <map>
#include <vector>
using namespace std;
const int MAXN=250010;
int dp[MAXN];

int nValue;
//0-1背包，代价为cost,获得的价值为weight
void ZeroOnePack(int cost,int weight)
{
    for(int i=nValue;i>=cost;i--)
        dp[i]=max(dp[i],dp[i-cost]+weight);
}
//完全背包，代价为cost,获得的价值为weight
void CompletePack(int cost,int weight)
{
    for(int i=cost;i<=nValue;i++)
        dp[i]=max(dp[i],dp[i-cost]+weight);
}
//多重背包
void MultiplePack(int cost,int weight,int amount)
{
    if(cost*amount>=nValue)CompletePack(cost,weight);
    else
    {
        int k=1;
        while(k<amount)
        {
            ZeroOnePack(k*cost,k*weight);
            amount-=k;
            k<<=1;
        }
        ZeroOnePack(amount*cost,amount*weight);
    }
}
int a[55],b[55];
int main()
{
    int n;
    while(scanf("%d",&n)==1)
    {
        if(n<0)break;
        nValue=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&a[i],&b[i]);
            nValue+=a[i]*b[i];
        }
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;i++)
        {
            if(b[i]==1)ZeroOnePack(a[i],a[i]);
            else MultiplePack(a[i],a[i],b[i]);
        }
        int t=nValue/2;
        while(dp[t]!=t)t--;
        printf("%d %d\n",nValue-t,t);
    }
    return 0;
}