POJ 1014 Dividing 解题报告

一、问题描述

 

二、解答思路

使用动态规划。定义一个数组b[60005]。b[i]表示是否可以分成i和sum-i两份。显然b[0]=true。如果b[sum/2]==true，则说明可以分割。如果b[k]=true，则b[k+i]=true，i为石子价值。

 

三、代码

 

#include<iostream>
#include<string.h>
using namespace std;
int num[7];
bool b[60005];

int main()
{
    int i,j,k;
    int sum ;
    int c=1;
    bool first=true;
    int half;
    bool flag;

    while(scanf("%d%d%d%d%d%d",&num[1],&num[2],&num[3],&num[4],&num[5],&num[6])!=EOF)
    {
        if(num[1]==0 && num[2]==0 && num[3]==0 && num[4]==0 && num[5]==0 && num[6]==0)
            break;

        //for(i=1;i<=6;++i)

            //num[i]=num[i]%60;

        flag=false;
        memset(b,0,sizeof(b));
        b[0]=true;
        if(first ==false)
        {
            printf("\n");
        }
        else
            first=false;
        printf("Collection #%d:\n",c);
        c++;
        sum=0;
        for(i=1;i<=6;++i)
            sum+=(i*num[i]);
        if(sum % 2 ==1)
        {
            printf("Can't be divided.\n");
            continue;
        }
        half=sum/2;
        int Max=0;
        for(i=1;i<=6;++i)
        {
            if(b[half])
                break;
            for(j=1;j<=num[i];++j)
            {
                if(b[half-i])
                {
                    b[half]=true;
                    printf("Can be divided.\n");
                    break;
                }
                for(k=Max;k>=0;k--)
                {
                    if(b[k] && k+i<=half)
                    {
                        if(b[k+i])
                            break;
                        else
                            b[k+i]=true;
                        if(k+i > Max)
                            Max=k+i;
                    }
                }
            }
        }
        if(b[half]==false)
            printf("Can't be divided.\n");
    }
    return 0;
}