ZOJ 1149（多重背包）

最新推荐文章于 2022-05-03 19:30:00 发布

墨温温

最新推荐文章于 2022-05-03 19:30:00 发布

阅读量418

点赞数

CC 4.0 BY-SA版权

分类专栏：背包 ZOJ 文章标签： ACM zoj

本文链接：https://blog.youkuaiyun.com/mowenwen_/article/details/48250523

背包同时被 2 个专栏收录

3 篇文章

订阅专栏

ZOJ

1 篇文章

订阅专栏

解决Marsha和Bill如何公平地分配不同价值的大理石集合的问题。通过编程实现多重背包问题的算法来判断是否可以将大理石分为两组，使得每组的总价值相同。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Dividing

Time Limit: 2000MS

Memory Limit: 65536KB

64bit IO Format: %lld & %llu

Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the same value, because then they could just split the collection in half. But unfortunately, some of the marbles are larger, or more beautiful than others. So, Marsha and Bill start by assigning a value, a natural number between one and six, to each marble. Now they want to divide the marbles so that each of them gets the same total value.

Unfortunately, they realize that it might be impossible to divide the marbles in this way (even if the total value of all marbles is even). For example, if there are one marble of value 1, one of value 3 and two of value 4, then they cannot be split into sets of equal value. So, they ask you to write a program that checks whether there is a fair partition of the marbles.

Input

Each line in the input describes one collection of marbles to be divided. The lines consist of six non-negative integers n1, n2, ..., n6, where ni is the number of marbles of value i. So, the example from above would be described by the input-line ``1 0 1 2 0 0''. The maximum total number of marbles will be 20000.

The last line of the input file will be ``0 0 0 0 0 0''; do not process this line.

Output

For each colletcion, output ``Collection #k:'', where k is the number of the test case, and then either ``Can be divided.'' or ``Can't be divided.''.

Output a blank line after each test case.

Sample Input

1 0 1 2 0 0
1 0 0 0 1 1
0 0 0 0 0 0

Sample Output

Collection #1:
Can't be divided.

Collection #2:
Can be divided.

Source

Mid-Central European Regional Contest 1999

多重背包问题~
AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define inf 0x7fffffff
int a[7],dp[120010];
int sum , v;
//01背包
void ZeroOnePack(int cost,int weight,int lim)
{
    for(int i=lim;i>=cost;i--)
       dp[i]=max(dp[i],dp[i-cost]+weight);
}
//完全背包
void CompletePack(int cost,int weight,int lim)
{
    for(int i=cost;i<=lim;i++)
      dp[i]=max(dp[i],dp[i-cost]+weight);
}
//多重背包
void MultiplePack(int cost,int weight,int amount,int lim)
{
    if(cost*amount>=lim) CompletePack(cost,weight,lim);
    else
    {
        for(int k=1;k<amount;)
        {
            ZeroOnePack(k*cost,k*weight,lim);
            amount-=k;
            k<<=1;
        }
        ZeroOnePack(amount*cost,amount*weight,lim);
    }
}

int main()
{
    int cas = 1;
    while(scanf("%d%d%d%d%d%d",&a[1], &a[2], &a[3], &a[4], &a[5], &a[6]) != EOF)
    {
        if(a[1]+a[2]+a[3]+a[4]+a[5]+a[6] == 0)
            break;
        printf("Collection #%d:\n",cas++);
        sum = a[1]+2*a[2]+3*a[3]+4*a[4]+5*a[5]+6*a[6];
        if(sum & 1)
            printf("Can't be divided.\n");
        else
        {
            memset(dp, 0, sizeof(dp));
            v = sum / 2;
            for(int i = 1;i <= 6;i ++)
                MultiplePack(i,i,a[i],v);
            if(dp[v] == v)
                printf("Can be divided.\n");
            else
                printf("Can't be divided.\n");
        }
        printf("\n");
    }
    return 0;
}