bzoj 1054 bfs

　　就是bfs，对于每个状态存一个hash为当前状态矩阵的二进制表示，然后搜就行了，写成双向bfs会快很多。

　　反思：对于C++的数组从0开始还不是特别习惯，经常犯错，对于C++的结构体不熟。

/**************************************************************
    Problem: 1054
    User: BLADEVIL
    Language: C++
    Result: Accepted
    Time:112 ms
    Memory:1940 kb
****************************************************************/

//By BLADEVIL
#include <cstdio>
#include <iostream>
#include <queue>
#include <set>
#define LL long long

using namespace std;

const int go[][]={{-,},{,},{,},{,-}};

struct rec
{
    int map[][];
};

struct node
{
    rec x;
    int y;
    node(rec xx,int yy):x(xx),y(yy){}
};

LL hash(rec x)
{
    LL h=;
    for (int i=;i<=;i++)
        for (int j=;j<=;j++)
            h^=(x.map[i][j]<<(*(i-)+(j-)));
    return h;
}

int main()
{
    queue<node>que;
    set<LL>bt;
    rec start,finish;
    for (int i=;i<=;i++)
    {
        char c[];
        scanf("%s",&c);
        for (int j=;j<;j++) start.map[i][j+]=(c[j]=='')?:;
    }
    for (int i=;i<=;i++)
    {
        char c[];
        scanf("%s",&c);
        for (int j=;j<;j++) finish.map[i][j+]=(c[j]=='')?:;
    }
    //printf("%lld %lld\n",hash(start),hash(finish));
    if (hash(start)==hash(finish))
    {
        printf("0\n");
        return ;
    }
    que.push(node(start,)); bt.insert(hash(start));
    while (!que.empty())
    {
        node cur=que.front(); que.pop();
        //printf("%lld ",hash(cur.x));
        for (int i=;i<=;i++)
            for (int j=;j<=;j++) if (cur.x.map[i][j])
                for (int k=;k<;k++)
                {
                    int x=i+go[k][],y=j+go[k][];
                    //printf("%d %d ",i,j);
                    //printf("%d %d |",x,y);
                    if ((!x)||(x>)||(!y)||(y>)) continue;
                    if (cur.x.map[x][y]) continue;
                    rec next=cur.x;
                    next.map[x][y]=next.map[i][j]--;
                    if (bt.count(hash(next))) continue;
                    que.push(node(next,cur.y+)); bt.insert(hash(next));
                    if (hash(finish)==hash(next))
                    {
                        printf("%d\n",cur.y+);
                        return ;
                    }
                }
    }
    return ;
}