hdu1495(经典bfs，平分水问题)

思路：搜索题，第一次做这种类型的题目吧，一开始表示不怎么明白题意所说的东东。其实就是要你判断可乐能不能被平分........

有六种状态，从a瓶到b瓶，a-->c

b-->a     b-->c

c-->a     c-->b

然后每种状态里面又分两种不同情况，可以将此瓶的水全部清空，不能清空......

然后广搜就可以了........

#include<iostream>

#include<stdio.h>

#include<string.h>

#include<queue>

using namespace std;

int vist[105][105][105],a,b,c;

struct node

{

    int a,b,c;

    int step;

}s[105];

int sum=0;

void bfs()

{

    queue<node>q;

    memset(vist,0,sizeof(vist));

    node p1;

    p1.a=a;

    p1.b=0;

    p1.c=0;

    p1.step=0;

    q.push(p1);

    vist[p1.a][0][0]=1;

    while(!q.empty())

    {

        p1=q.front();

        q.pop();

        if((p1.a==a/2&&p1.b==a/2)||(p1.a==a/2&&p1.c==a/2)||(p1.b==a/2&&p1.c==a/2))

        {

            printf("%d\n",p1.step);

            return;

        }

        node p2;    

        if(p1.a!=0)

        {

            if(p1.a>b-p1.b)

            {

                p2.a=p1.a-(b-p1.b);

                p2.b=b;

                p2.c=p1.c;

                p2.step=p1.step+1;

            }

            else

            {

                p2.a=0;

                p2.b=p1.b+p1.a;

                p2.c=p1.c;

                p2.step=p1.step+1;

            }

            if(!vist[p2.a][p2.b][p2.c])

            {

                vist[p2.a][p2.b][p2.c]=1;

                q.push(p2);

            }

        }

        

        if(p1.a!=0)

        {

            if(p1.a>c-p1.c)

            {

                p2.a=p1.a-(c-p1.c);

                p2.b=p1.b;

                p2.c=c;

                p2.step=p1.step+1;

            }

            else

            {

                p2.a=0;

                p2.b=p1.b;

                p2.c=p1.c+p1.a;

                p2.step=p1.step+1;

            }

            if(!vist[p2.a][p2.b][p2.c])

            {

                vist[p2.a][p2.b][p2.c]=1;

                q.push(p2);

            }

        }

        

        if(p1.b!=0)

        {

            if(p1.b>a-p1.a)

            {

                p2.b=p1.b-(a-p1.a);

                p2.a=a;

                p2.c=p1.c;

                p2.step=p1.step+1;

            }

            else

            {

                p2.b=0;

                p2.a=p1.a+p1.b;

                p2.c=p1.c;

                p2.step=p1.step+1;

            }

            if(!vist[p2.a][p2.b][p2.c])

            {

                vist[p2.a][p2.b][p2.c]=1;

                q.push(p2);

            }

        }

        

        if(p1.b!=0)

        {

            if(p1.b>c-p1.c)

            {

                p2.b=p1.b-(c-p1.c);

                p2.a=p1.a;

                p2.c=c;

                p2.step=p1.step+1;

            }

            else

            {

                p2.b=0;

                p2.a=p1.a;

                p2.c=p1.c+p1.b;

                p2.step=p1.step+1;

            }

            if(!vist[p2.a][p2.b][p2.c])

            {

                vist[p2.a][p2.b][p2.c]=1;

                q.push(p2);

            }

        }

        

        if(p1.c!=0)

        {

            if(p1.c>a-p1.a)

            {

                p2.c=p1.c-(a-p1.a);

                p2.a=a;

                p2.b=p1.b;

                p2.step=p1.step+1;

            }

            else

            {

                p2.c=0;

                p2.a=p1.a+p1.c;

                p2.b=p1.b;

                p2.step=p1.step+1;

            }

            if(!vist[p2.a][p2.b][p2.c])

            {

                vist[p2.a][p2.b][p2.c]=1;

                q.push(p2);

            }

        }

        

        if(p1.c!=0)

        {

            if(p1.c>b-p1.b)

            {

                p2.c=p1.c-(b-p1.b);

                p2.a=p1.a;

                p2.b=b;

                p2.step=p1.step+1;

            }

            else

            {

                p2.c=0;

                p2.a=p1.a;

                p2.b=p1.b+p1.c;

                p2.step=p1.step+1;

            }

            if(!vist[p2.a][p2.b][p2.c])

            {

                vist[p2.a][p2.b][p2.c]=1;

                q.push(p2);

            }

        }

    }

    printf("NO\n");

}

int main()

{

    while(scanf("%d%d%d",&a,&b,&c)>0&&(a+b+c))

    {

        if(a%2==1)

        {

            printf("NO\n");

            continue;

        }

        bfs();

    }

    return 0;

}