UVA 10651 - Pebble Solitaire

这道题是模仿别人的bfs，却也有所收获。把12个棋位的有无棋子整体看作一个状态，然后宽搜就可以了。

代码如下：

#include<stdio.h>
#include<string.h>
#define MAXN 11000
char b[20];
int q[MAXN], hash[MAXN], d[MAXN];
int N, n;
void solve()
{
    int x,i;
    int front = 0, rear = 0;
    memset(hash,0,sizeof(hash));
    hash[q[rear]] = 1;
    d[rear] = 0;
    rear ++;
    while(front < rear)
    {
        x = q[front];
        for(i = 0; i < 12; i ++)
            if((1 << i) & x)
            {
                if(i < 10 && ((1 << (i+1)) & x) && !((1 << (i+2)) & x))
                {
                    q[rear] = x ^ (1 << i) ^ (1 << (i+1)) ^ (1 << (i+2));
                    if(!hash[q[rear]])
                    {
                        d[rear] = d[front] + 1;
                        hash[q[rear]] = 1;
                        rear ++ ;
                    }
                }
                if(i > 1 && ((1 << (i-1)) & x) && !((1 << (i-2)) & x))
                {
                    q[rear] = x ^ (1 << i) ^ (1 << (i-1)) ^ (1 << (i-2));
                    if(!hash[q[rear]])
                    {
                        d[rear] = d[front] + 1;
                        hash[q[rear]] = 1;
                        rear ++;
                    }
                }
            }
        front ++;
    }
    printf("%d\n",N - d[rear - 1]);
}
void input()
{
    while(scanf("%d",&n) == 1)
    {
        while(n --)
        {
            scanf("%s",b);
            N = q[0] = 0;
            for(int i = 0; i < 12; i ++)
                q[0] = (q[0]<<1) + (b[i] == 'o' ? (N ++,1) : 0);
            solve();        
        }
    }
}
int main()
{
    input();
    return 0;
}