BZOJ1054 [HAOI2008]移动玩具

有时候题解要写的简略一点...于是：

裸的bfs妥妥的。。。

注意hash函数的选取及判断重复即可

结果memcpy函数写错了，调了好长好长时间、、、哭T T...

（据说不判重也可以过。。。汗！）

/**************************************************************
    Problem: 1054
    User: rausen
    Language: C++
    Result: Accepted
    Time:28 ms
    Memory:3060 kb
****************************************************************/
 
#include <cstdio>
#include <cstring>
#include <algorithm>
  
using namespace std;
const int dx[4] = {0, 0, 1, -1};
const int dy[4] = {1, -1, 0, 0};
bool Ans[5][5], vis[70000];
struct data{
    bool a[5][5];
    int step;
}q[70000];
int begin, end, H;
int l, r;
  
int hash(bool a[5][5]){
    int k = 1, res = 0, i, j;
    for (i = 1; i <= 4; ++i)
        for (j = 1; j <= 4; ++j)
            res += k * a[i][j], k <<= 1;
    return res;
}
  
int main(){
    char ch[5];
    int i, j, k, X, Y;
    for (i = 1; i <= 4; ++i)
        for (j = 1, scanf("%s", ch); j <= 4; ++j)
            q[0].a[i][j] = ch[j - 1] - '0';
    for (i = 1; i <= 4; ++i)     
        for (j = 1, scanf("%s", ch); j <= 4; ++j)
            Ans[i][j] = ch[j - 1] - '0';
    begin = hash(q[0].a);
    end = hash(Ans);
    if (begin == end){
        printf("0
");
        return 0;
    }
    vis[begin] = 1;
    for (l = r = 0; l <= r; ++l)
        for (i = 1; i <= 4; ++i)
            for (j = 1; j <= 4; ++j)
                if (q[l].a[i][j]) for (k = 0; k < 4; ++k){
                    X = i + dx[k], Y = j + dy[k];
                    if (X < 1 || Y < 1 || X > 4 || Y > 4 || q[l].a[X][Y])
                        continue;
                    swap(q[l].a[i][j], q[l].a[X][Y]);
                    H = hash(q[l].a);
                    if (!vis[H]){
                        if (H == end){
                            printf("%d
", q[l].step + 1);
                            return 0;
                        }
                        vis[H] = 1;
                        q[++r].step = q[l].step + 1;    
                        memcpy(q[r].a, q[l].a, sizeof(q[r].a));
                    }
                    swap(q[l].a[i][j], q[l].a[X][Y]);
                }
    return 0;
}