非常可乐(bfs)

C - 非常可乐

Time Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1495

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 

AC:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>

using namespace std;

struct Node
{
    int a, b, c, step;
};
int vis[110][110][110];
queue <Node> q;

int main()
{
    int s, n, m, flag;
    Node n1, n2;
    while(scanf("%d%d%d", &s, &n, &m) && s && n && m )
    {
        while( !q.empty() )
        {
            q.pop();
        }
        flag = 0;
        memset(vis, 0, sizeof(vis));
        if( s%2 != 0 )
        {
            printf("NO\n");
            continue;
        }
        n1.a = s;
        n1.b = 0;
        n1.c = 0;
        n1.step = 0;
        q.push(n1);
        vis[n1.a][n1.b][n1.c] = 1;
        while( !q.empty() )
        {
            n1 = q.front();
            q.pop();
            if( (n1.a == s/2 && n1.b == s/2) || (n1.c == s/2 && n1.b == s/2) || (n1.a == s/2 && n1.c == s/2) )
            {
                flag = 1;
                break;
            }
            if( n1.a )
            {
                if( n1.b<n )
                {
                    n2.a = (n-n1.b)>n1.a?0:(n1.a-(n-n1.b));
                    n2.b = (n-n1.b)>n1.a?(n1.a+n1.b):n;
                    n2.c = n1.c;
                    if( !vis[n2.a][n2.b][n2.c] )
                    {
                        vis[n2.a][n2.b][n2.c] = 1;
                        n2.step = n1.step+1;
                        q.push(n2);
                    }
                }
                if( n1.c<m )
                {
                    n2.a = (m-n1.c)>n1.a?0:(n1.a-(m-n1.c));
                    n2.c = (m-n1.c)>n1.a?(n1.a+n1.c):m;
                    n2.b = n1.b;
                    if( !vis[n2.a][n2.b][n2.c] )
                    {
                        vis[n2.a][n2.b][n2.c] = 1;
                        n2.step = n1.step+1;
                        q.push(n2);
                    }
                }
            }
            if( n1.b )
            {
                if( n1.a<s )
                {
                    n2.b = (s-n1.a)>n1.b?0:(n1.b-(s-n1.a));
                    n2.a = (s-n1.a)>n1.b?(n1.a+n1.b):s;
                    n2.c = n1.c;
                    if( !vis[n2.a][n2.b][n2.c] )
                    {
                        vis[n2.a][n2.b][n2.c] = 1;
                        n2.step = n1.step+1;
                        q.push(n2);
                    }
                }
                if( n1.c<m )
                {
                    n2.b = (m-n1.c)>n1.b?0:(n1.b-(m-n1.c));
                    n2.c = (m-n1.c)>n1.b?(n1.b+n1.c):m;
                    n2.a = n1.a;
                    if( !vis[n2.a][n2.b][n2.c] )
                    {
                        vis[n2.a][n2.b][n2.c] = 1;
                        n2.step = n1.step+1;
                        q.push(n2);
                    }
                }
            }
            if( n1.c )
            {
                if( n1.a<s )
                {
                    n2.c = (s-n1.a)>n1.c?0:(n1.c-(s-n1.a));
                    n2.a = (s-n1.a)>n1.c?(n1.a+n1.c):s;
                    n2.b = n1.b;
                    if( !vis[n2.a][n2.b][n2.c] )
                    {
                        vis[n2.a][n2.b][n2.c] = 1;
                        n2.step = n1.step+1;
                        q.push(n2);
                    }
                }
                if( n1.b<n )
                {
                    n2.c = (n-n1.b)>n1.c?0:(n1.c-(n-n1.b));
                    n2.b = (n-n1.b)>n1.c?(n1.c+n1.b):n;
                    n2.a = n1.a;
                    if( !vis[n2.a][n2.b][n2.c] )
                    {
                        vis[n2.a][n2.b][n2.c] = 1;
                        n2.step = n1.step+1;
                        q.push(n2);
                    }
                }
            }
        }
        if( flag )
            printf("%d\n", n1.step);
        else
            printf("NO\n");
    }
    return 0;
}

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