poj 1014(多重背包)

最新推荐文章于 2020-01-21 17:44:41 发布

原创最新推荐文章于 2020-01-21 17:44:41 发布 · 281 阅读

1 ·

CC 4.0 BY-SA版权

acm 同时被 2 个专栏收录

71 篇文章

订阅专栏

27 篇文章

订阅专栏

本文深入探讨了游戏开发领域的关键技术，包括游戏引擎、编程语言、硬件优化等，并重点阐述了AI音视频处理的应用场景和实现方法，如语义识别、语音识别、AR增强现实等。

Dividing

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 66477		Accepted: 17291

Description

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 file describes one collection of marbles to be divided. The lines contain six non-negative integers n1 , . . . , 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 collection, 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

首先奇数肯定不能平分，偶数可以多重背包总和的一半看看最多能否取到一半

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
int dp[200005];
int a[10];
int val;
void ZeroOnePack(int w,int v)
{
    for(int i=val;i>=w;i--)
        dp[i]=max(dp[i],dp[i-w]+v);
}

void CompletePack(int w,int v)
{
    for(int i=w;i<=val;i++)
        dp[i]=max(dp[i],dp[i-w]+v);
}

int MultiplePack()
{
    for(int i=1;i<=6;i++)
    {
        if(a[i]*i>=val)
            CompletePack(i,i);
        else
        {
            int k=1;
            while(k<a[i])
            {
                ZeroOnePack(i*k,i*k);
                a[i]-=k;
                k*=2;
            }
            ZeroOnePack(i*a[i],a[i]*i);
        }
    }
    return dp[val];
}

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