HDU 1495 非常可乐（广搜）

于 2017-06-10 01:43:06 发布
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分类专栏： 搜索
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本文链接：https://blog.youkuaiyun.com/xuejye/article/details/72973655
本文介绍了一个有趣的可乐分配问题，并通过广度优先搜索算法找到了解决该问题的方法。问题要求将一定量的可乐平均分配到两个不同容量的杯子中，文章详细展示了如何使用结构体存储每一步的状态，以及如何利用队列实现广度优先搜索算法来寻找最少的操作步骤。

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1495点击打开链接

非常可乐

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14910    Accepted Submission(s): 5971

Problem Description
大家一定觉的运动以后喝可乐是一件很惬意的事情，但是seeyou却不这么认为。因为每次当seeyou买了可乐以后，阿牛就要求和seeyou一起分享这一瓶可乐，而且一定要喝的和seeyou一样多。但seeyou的手中只有两个杯子，它们的容量分别是N 毫升和M 毫升 可乐的体积为S （S<101）毫升　(正好装满一瓶) ，它们三个之间可以相互倒可乐 (都是没有刻度的，且 S==N+M，101＞S＞0，N＞0，M＞0) 。聪明的ACMER你们说他们能平分吗？如果能请输出倒可乐的最少的次数，如果不能输出"NO"。
 

Input
三个整数 : S 可乐的体积 , N 和 M是两个杯子的容量，以"0 0 0"结束。
 

Output
如果能平分的话请输出最少要倒的次数，否则输出"NO"。
 

Sample Input 
  

   7 4 3
4 1 3
0 0 0
 

Sample Output 
  

   NO
3
 

Author
seeyou

广搜 同时注意标记 因为两个状态确定 第三个状态随之也确定 因此用二维数组就能储存 更加简便

代码有些冗长 分情况讨论 

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include<algorithm>
#include <math.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <queue>
#include <stack>
using namespace std;
struct xjy
{
    int i;
    int j;
    int k;
    int number;
};
int b[4];
int book[111][111];
int n;int m;int s;
int num=0;
void bfs()
{
    book[s][0]=1;
    int q,w,e;
    q=s;w=0;e=0;
    xjy mid;xjy midmid;
    mid.i=q;mid.j=w;mid.k=e;mid.number=0;
    queue <xjy> qu;
    qu.push(mid);
    while(!qu.empty())
        {
            mid=qu.front();
            qu.pop();
            if(((mid.i+mid.j)==s&&mid.i==mid.j)||((mid.i+mid.k)==s&&mid.i==mid.k)||((mid.k+mid.j)==s&&mid.k==mid.j))
            {
                
                num=mid.number;
                break;
            };
            
            if(mid.i>=(b[2]-mid.j))
            {
                midmid=mid;
                midmid.i=midmid.i-b[2]+midmid.j;
                midmid.j=b[2];
                midmid.number++;
                if(book[midmid.i][midmid.j]!=1)
                {
                    qu.push(midmid);
                    book[midmid.i][midmid.j]=1;
                }
            }
            if(mid.i<(b[2]-mid.j))
            {
                midmid=mid;
                midmid.j+=midmid.i;
                midmid.i=0;
                midmid.number++;
                if(book[midmid.i][midmid.j]!=1)
                {
                    qu.push(midmid);
                    book[midmid.i][midmid.j]=1;
                }
            }
            
            if(mid.i>=(b[3]-mid.k))
            {
                midmid=mid;
                midmid.i=midmid.i-b[3]+midmid.k;
                midmid.k=b[3];
                midmid.number++;
                if(book[midmid.i][midmid.j]!=1)
                {
                    qu.push(midmid);
                    book[midmid.i][midmid.j]=1;
                }
            }
            if(mid.i<(b[3]-mid.k))
            {
                midmid=mid;
                midmid.k+=midmid.i;
                midmid.i=0;
                midmid.number++;
                if(book[midmid.i][midmid.j]!=1)
                {
                    qu.push(midmid);
                    book[midmid.i][midmid.j]=1;
                }
            }
            
            if(mid.j>=(b[3]-mid.k))
            {
                midmid=mid;
                midmid.j=midmid.j-b[3]+midmid.k;
                midmid.k=b[3];
                midmid.number++;
                if(book[midmid.i][midmid.j]!=1)
                {
                    qu.push(midmid);
                    book[midmid.i][midmid.j]=1;
                }
            }
            if(mid.j<(b[3]-mid.k))
            {
                midmid=mid;
                midmid.k+=midmid.j;
                midmid.j=0;
                midmid.number++;
                if(book[midmid.i][midmid.j]!=1)
                {
                    qu.push(midmid);
                    book[midmid.i][midmid.j]=1;
                }
            }
            
            if(mid.j>=(b[1]-mid.i))
            {
                midmid=mid;
                midmid.j=midmid.j-b[1]+midmid.i;
                midmid.i=b[1];
                midmid.number++;
                if(book[midmid.i][midmid.j]!=1)
                {
                    qu.push(midmid);
                    book[midmid.i][midmid.j]=1;
                }
            }
            if(mid.j<(b[1]-mid.i))
            {
                midmid=mid;
                midmid.i+=midmid.j;
                midmid.j=0;
                midmid.number++;
                if(book[midmid.i][midmid.j]!=1)
                {
                    qu.push(midmid);
                    book[midmid.i][midmid.j]=1;
                }
            }
            
            if(mid.k>=(b[1]-mid.i))
            {
                midmid=mid;
                midmid.k=midmid.k-b[1]+midmid.i;
                midmid.i=b[1];
                midmid.number++;
                if(book[midmid.i][midmid.j]!=1)
                {
                    qu.push(midmid);
                    book[midmid.i][midmid.j]=1;
                }
            }
            if(mid.k<(b[1]-mid.i))
            {
                midmid=mid;
                midmid.i+=midmid.k;
                midmid.k=0;
                midmid.number++;
                if(book[midmid.i][midmid.j]!=1)
                {
                    qu.push(midmid);
                    book[midmid.i][midmid.j]=1;
                }
            }
            
            if(mid.k>=(b[2]-mid.j))
            {
                midmid=mid;
                midmid.k=midmid.k-b[2]+midmid.j;
                midmid.j=b[2];
                midmid.number++;
                if(book[midmid.i][midmid.j]!=1)
                {
                    qu.push(midmid);
                    book[midmid.i][midmid.j]=1;
                }
            }
            if(mid.k<(b[2]-mid.j))
            {
                midmid=mid;
                midmid.j+=midmid.k;
                midmid.k=0;
                midmid.number++;
                if(book[midmid.i][midmid.j]!=1)
                {
                    qu.push(midmid);
                    book[midmid.i][midmid.j]=1;
                }
            }
        }
    }

int main()
{
    
    while(scanf("%d%d%d",&s,&n,&m)&&s&&n&&m)
    {
        if(n<m)
        {
            int t=n;
            n=m;
            m=t;
        }
        num=0;
        b[1]=s;
        b[2]=n;
        b[3]=m;
        for(int i=0;i<111;i++)
            for(int j=0;j<111;j++)
                    book[i][j]=0;
        bfs();
        if(num)
            printf("%d\n",num);
        else
            printf("NO\n");
    }
}

这次比赛又碰到了 附一下这次的吧 代码差不多

#include <iostream>
#include <queue>
#include <stdio.h>
#include <stdlib.h>
#include <stack>
#include <limits.h>
#include <string>
#include <string.h>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
using namespace std;
char mmap[11][11];
int book[111][111][111];
struct xjy
{
    int nv;
    int mv;
    int kv;
};
int ans=INT_MAX;
xjy b,e;
int n,m,k;
int num;
int cnt;
queue <xjy > q;
int dir[4][2]={1,0,-1,0,0,1,0,-1};
void bfs()
{
    while(!q.empty())
        q.pop();
    xjy mid;
    mid.nv=n;
    mid.mv=0;
    mid.kv=0;
    q.push(mid);
    while(!q.empty())
    {
        mid=q.front();
        q.pop();
        if(((mid.nv==mid.mv)&&mid.nv==n/2)||((mid.kv==mid.mv)&&mid.mv==n/2)||((mid.nv==mid.kv)&&mid.nv==n/2))
           {
               ans=book[mid.nv][mid.mv][mid.kv];
               break;
           }
//cout << mid.nv<<" "<< mid.mv << " "<< mid.kv << endl;
        xjy midmid;
        midmid=mid;
        if(midmid.nv!=0||midmid.mv!=m)
        {
            if(midmid.nv>(m-midmid.mv))
            {
                midmid.nv-=(m-midmid.mv);
                midmid.mv=m;
                if(!book[midmid.nv][midmid.mv][midmid.kv])
                {
                    q.push(midmid);
                    book[midmid.nv][midmid.mv][midmid.kv]=book[mid.nv][mid.mv][mid.kv]+1;
                }
            }
            else
            {
                midmid.mv+=midmid.nv;
                midmid.nv=0;
                if(!book[midmid.nv][midmid.mv][midmid.kv])
                {
                    q.push(midmid);
                    book[midmid.nv][midmid.mv][midmid.kv]=book[mid.nv][mid.mv][mid.kv]+1;
                }
            }
        }

        midmid=mid;
        if(midmid.nv!=0||midmid.kv!=k)
        {
            if(midmid.nv>(k-midmid.kv))
            {
                midmid.nv-=(k-midmid.kv);
                midmid.kv=k;
                if(!book[midmid.nv][midmid.mv][midmid.kv])
                {
                    q.push(midmid);
                    book[midmid.nv][midmid.mv][midmid.kv]=book[mid.nv][mid.mv][mid.kv]+1;
                }
            }
            else
            {
                midmid.kv+=midmid.nv;
                midmid.nv=0;
                if(!book[midmid.nv][midmid.mv][midmid.kv])
                {
                    q.push(midmid);
                    book[midmid.nv][midmid.mv][midmid.kv]=book[mid.nv][mid.mv][mid.kv]+1;
                }
            }
        }

        midmid=mid;
        if(midmid.mv!=0||midmid.kv!=k)
        {
            if(midmid.mv>(k-midmid.kv))
            {
                midmid.mv-=(k-midmid.kv);
                midmid.kv=k;
                if(!book[midmid.nv][midmid.mv][midmid.kv])
                {
                    q.push(midmid);
                    book[midmid.nv][midmid.mv][midmid.kv]=book[mid.nv][mid.mv][mid.kv]+1;
                }
            }
            else
            {
                midmid.kv+=midmid.mv;
                midmid.mv=0;
                if(!book[midmid.nv][midmid.mv][midmid.kv])
                {
                    q.push(midmid);
                    book[midmid.nv][midmid.mv][midmid.kv]=book[mid.nv][mid.mv][mid.kv]+1;
                }
            }
        }

        midmid=mid;
        if(midmid.mv!=0||midmid.nv!=n)
        {
            if(midmid.mv>(n-midmid.nv))
            {
                midmid.mv-=(n-midmid.nv);
                midmid.nv=n;
                if(!book[midmid.nv][midmid.mv][midmid.kv])
                {
                    q.push(midmid);
                    book[midmid.nv][midmid.mv][midmid.kv]=book[mid.nv][mid.mv][mid.kv]+1;
                }
            }
            else
            {
                midmid.nv+=midmid.mv;
                midmid.mv=0;
                if(!book[midmid.nv][midmid.mv][midmid.kv])
                {
                    q.push(midmid);
                    book[midmid.nv][midmid.mv][midmid.kv]=book[mid.nv][mid.mv][mid.kv]+1;
                }
            }
        }

        midmid=mid;
        if(midmid.kv!=0||midmid.mv!=m)
        {
            if(midmid.kv>(m-midmid.mv))
            {
                midmid.kv-=(m-midmid.mv);
                midmid.mv=m;
                if(!book[midmid.nv][midmid.mv][midmid.kv])
                {
                    q.push(midmid);
                    book[midmid.nv][midmid.mv][midmid.kv]=book[mid.nv][mid.mv][mid.kv]+1;
                }
            }
            else
            {
                midmid.mv+=midmid.kv;
                midmid.kv=0;
                if(!book[midmid.nv][midmid.mv][midmid.kv])
                {
                    q.push(midmid);
                    book[midmid.nv][midmid.mv][midmid.kv]=book[mid.nv][mid.mv][mid.kv]+1;
                }
            }
        }

        midmid=mid;
        if(midmid.kv!=0||midmid.nv!=n)
        {
            if(midmid.kv>(n-midmid.nv))
            {
                midmid.kv-=(n-midmid.nv);
                midmid.nv=n;
                if(!book[midmid.nv][midmid.mv][midmid.kv])
                {
                    q.push(midmid);
                    book[midmid.nv][midmid.mv][midmid.kv]=book[mid.nv][mid.mv][mid.kv]+1;
                }
            }
            else
            {
                midmid.nv+=midmid.kv;
                midmid.kv=0;
                if(!book[midmid.nv][midmid.mv][midmid.kv])
                {
                    q.push(midmid);
                    book[midmid.nv][midmid.mv][midmid.kv]=book[mid.nv][mid.mv][mid.kv]+1;
                }
            }
        }

    }
}

int main()
{

    while(~scanf("%d%d%d",&n,&m,&k)&&(n||m||k))
    {
        memset(book,0,sizeof(book));
        ans=INT_MAX;
        book[n][0][0]=1;
        if(n&1)
        {
            printf("NO\n");
            continue;
        }
        else
        {

        bfs();
        if(ans!=INT_MAX)
            printf("%d\n",ans-1);
        else
            printf("NO\n");
        }
    }
}

 


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