非常可乐（杭电hdu1495）bfs

非常可乐 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3510    Accepted Submission(s): 1440 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

 这是一题搜索题，bfs，我的思路是分为A->B , A->C , B->A , B->C , C->B , C->A 六种情况，后面4种情况，要考虑是否溢出，，，仔细一点就不会错了，哈哈

附上我的代码，有什么更好的方法，请多多指教，                                                            ps：原题 http://acm.hdu.edu.cn/showproblem.php?pid=1495

#include<iostream>

#include<queue>

#include<cstring>

using namespace std;

int visit[105][105][105];

int s,m,n;



struct node

{

    int x,y,z;

    int step;

};



void bfs()

{

    queue<node>Q;    

    node now,next;

    now.x=s;

    now.y=0;

    now.z=0;

    visit[now.x][now.y][now.z]=1;

    now.step=0;

    Q.push(now);

    while(!Q.empty())

    {

        now=Q.front();

        Q.pop();

        if((now.x==s/2&&now.x==now.y)||(now.x==s/2&&now.x==now.z)||(now.y==s/2&&now.y==now.z))//到达评分条件的时候跳出

        {

            printf("%d\n",now.step);

            return ;

        }

        if(now.x != 0) //A->B,A->C

        {

            if(now.y != n)

            {

                next.x = now.x - (n - now.y);

                next.y = n;

                next.z = now.z;

    

                if(!visit[next.x][next.y][next.z])

                {

                    next.step = now.step + 1;

                    Q.push(next);

                    visit[next.x][next.y][next.z] = 1;

                }

            }

            if(now.z != m)

            {

                next.x = now.x - (m - now.z);

                next.y = now.y;

                next.z = m;



                if(!visit[next.x][next.y][next.z])

                {

                    next.step = now.step + 1;

                    Q.push(next);

                    visit[next.x][next.y][next.z] = 1;

                }

            }

        }

        if(now.y != 0) //B->A,B->C

        {

            next.x = now.x + now.y;

            next.y = 0;

            next.z = now.z;

  

            if(!visit[next.x][next.y][next.z])

            {

                next.step = now.step + 1;

                Q.push(next);

                visit[next.x][next.y][next.z] = 1;

            }

            if(now.y < m - now.z)

            {

                next.x = now.x;

                next.y = 0;

                next.z = now.y + now.z;

                if(!visit[next.x][next.y][next.z])

                {

                    next.step = now.step + 1;

                    Q.push(next);

                    visit[next.x][next.y][next.z] = 1;

                }

            }

            else

            {

                next.x = now.x;

                next.y = now.y - (m - now.z);

                next.z = m;

                if(!visit[next.x][next.y][next.z])

                {

                    next.step = now.step + 1;

                    Q.push(next);

                    visit[next.x][next.y][next.z] = 1;

                }

            }

        }

        if(now.z != 0) //C->A,C->B

        {

            next.x = now.x + now.z;

            next.y = now.y;

            next.z = 0;

            if(!visit[next.x][next.y][next.z])

                {

                    next.step = now.step + 1;

                    Q.push(next);

                    visit[next.x][next.y][next.z] = 1;

                }

            if(now.z < n - now.y)

            {

                next.x = now.x;

                next.y = now.y + now.z;

                next.z = 0;

                if(!visit[next.x][next.y][next.z])

                {

                    next.step = now.step + 1;

                    Q.push(next);

                    visit[next.x][next.y][next.z] = 1;

                }

            }

            else

            {

                next.x = now.x;

                next.y = n;

                next.z = now.z - (n - now.y);

                if(!visit[next.x][next.y][next.z])

                {

                    next.step = now.step + 1;

                    Q.push(next);

                    visit[next.x][next.y][next.z] = 1;

                }

            }

        }



    }

    printf("NO\n");

}

int main()

{

    while(scanf("%d%d%d",&s,&n,&m)!=EOF&&n||m||s)

    {

        memset(visit,0,sizeof(visit));

        if(s%2==0)//s为奇数是不可能平分的，可优化。

            bfs();

        else

            printf("NO\n");

    }

    return 0;

}

代码有点冗长啊，不过有很多是重复的，求更好的方法！欢迎评论