POJ之1014

来源：互联网发布：手机打不开数据流量编辑：程序博客网时间：2024/05/22 07:53

/*
两人分石头，要求平分，但石头大小不一，价值也不一样，从1到20000，如果一堆石头的总价是奇数显然不可分，则输出不可分割
否则再看这堆石头是否能被分成两份，在保证石头完好无损的情况下。
输入1 0 1 2 0 0代表价值为1的1个，价值为2的0个，·····；即有四块价值分别为4,4,3,1的石头，显然这石头是没法分的
输入1 0 0 0 1 1代表·····，即有三块价值为6,5,1的石头，显然可分，一份为6，另一份为5,1；
有两种解决方法：
第一种是同学都熟悉的多重背包问题
第二种方法是很少有同学使用的DFS

#include <iostream>#include<stdio.h>#include<string>#include<algorithm>#include<vector>#include<queue>#include<set>#include<map>#include<math.h>#include<limits.h>#include<stdlib.h>using namespace std;bool divided=false;int marble[7]= {0};void DFS(int value,int target,int pre){    if(divided)        return;    if(value==target)    {        divided=true;        return;    }    for(int i=pre; i>=1; i--)    {        if(marble[i])        {            if(value+i<=target)            {                marble[i]--;                DFS(value+i,target,i);                if(divided)                    break;            }        }    }    return;}int main(){    int pre = 6;    int evenValue_of_marble;//石块的均值    int flag = 1;    while(true)    {        int i=1;        int value_of_marble = 0;//所有石块的价值        int number_of_marble = 0;//石块的个数        while(i <= pre)        {            cin >> marble[i];            value_of_marble += marble[i] * i;            number_of_marble += marble[i];            i++;        }        if(number_of_marble > 20000)        {            break;        }        if(number_of_marble > 0)        {            if(value_of_marble % 2 == 0)            {                evenValue_of_marble = value_of_marble / 2;                DFS(0,evenValue_of_marble,pre);                if(divided)                {                    cout<<"Collection #"<<flag<<":"<<endl;                    cout<<"Can be divided."<<endl<<endl;                    flag++;                    continue;                }                else                {                    cout<<"Collection #"<<flag<<":"<<endl;                    cout<<"Can't be divided."<<endl<<endl;                    flag++;                    continue;                }            }            else            {                cout<<"Collection "<<"#"<<flag<<":"<<endl;                cout<<"Can't be divided."<<endl<<endl;                flag++;                continue;            }        }        else        {            break;        }    }    return 0;}

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