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
",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
");
}
int main()
{
    while(scanf("%d%d%d",&a,&b,&c)>0&&(a+b+c))
    {
        if(a%2==1)
        {
            printf("NO
");
            continue;
        }
        bfs();
    }
    return 0;
}