搜索的题

#include<cstdio>
#include<cstring>
#include<queue>
#include<iostream>
using namespace std;
const int hashn = 999997;
int hash[hashn];
struct node{
    int s[3][3];
};

struct node1{
    node point;
    int step;
    int x, y;
};

queue<node1> q;
node dest = {1,2,3,8,0,4,7,6,5};
const int dx[] = {0, 0, 1, -1};
const int dy[] = {-1, 1, 0, 0};

bool operator == (node &x, node &y){
    for(int i = 0; i < 3; i++)
        for(int j = 0; j < 3; j++)
            if(x.s[i][j] != y.s[i][j])    return false;
    return true;
}

void bfs(){
    while(!q.empty()){
        node1 u = q.front(); q.pop();
        for(int i = 0; i < 4; i++){
            int x1 = u.x + dx[i];
            int y1 = u.y + dy[i];
            if(x1 < 0 || x1 >= 3 || y1 < 0 || y1 >= 3)    continue;
            node now = u.point;
            now.s[u.x][u.y] = now.s[x1][y1];
            now.s[x1][y1] = 0;
            if(now == dest){
                cout << u.step + 1 << endl;
                return;
            }
            int n = 0;
            for(int j = 0; j < 3; j++)
                for(int k = 0; k < 3; k++)
                    n = n*10 + now.s[j][k];
            int nn = n % hashn;
            while(hash[nn] != n && nn < hashn){
                if(hash[nn] == -1)    break;
                nn++;
            }
            if(hash[nn] == -1){
                node1 v;
                hash[nn] = n;
                v.point = now;
                v.x = x1, v.y = y1;
                v.step = u.step + 1;
                q.push(v);
            }
        }
    }
}

int main(){
    memset(hash, -1, sizeof(hash));
    char c;
    node st;
    int x0, y0, n = 0;
    for(int i = 0; i < 3; i++)
        for(int j = 0; j < 3; j++){
             cin >> c;
             if(c == ‘0‘){
                 x0 = i; y0 = j;
             }
             st.s[i][j] = c - ‘0‘;
             n = n*10 + st.s[i][j];
        }
    int nn = n % hashn;
    hash[nn] = n;
    node1 st0;
    st0.point = st; st0.step = 0; st0.x = x0; st0.y = y0;
    q.push(st0);
    bfs();
    return 0;
}

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int xx[5] = {0,0,0,1,-1};
int yy[5] = {0,1,-1,0,0};

int map[5][5],minn = 1000;//利用数组代表“黑、白、空” minn代表所求的最小步数 
char s;
void Dfs(int x,int y,int num,int b)
{
    int m = 100000,i,j,k;
    if(num >= minn) 
        return;
    for(i = 1;i <= 4;i ++)
    {
        if(map[i][1] == map[i][2] && map[i][2] == map[i][3] && map[i][3] == map[i][4] && (map[i][4] == 1 || map[i][4] == 2))//行相同更新 
        m = num;
        if(map[1][i] == map[2][i] && map[2][i] == map[3][i] && map[3][i] == map[4][i] && (map[4][i] == 1 || map[4][i] == 2))//列相同更新 
        m = num;
    }
    if(map[1][1] == map[2][2] && map[2][2] == map[3][3] && map[3][3] == map[4][4] && (map[4][4]==1 || map[4][4] == 2))//左上到右下相同更新 
        m = num;
    if(map[4][1] == map[3][2] && map[3][2] == map[2][3] && map[2][3] == map[1][4] && (map[1][4]==1 || map[1][4] == 2))//右上到左下相同更新 
        m = num;
    if(m < minn)//m在上面已经更新为当前接了 这里用m更新minn 
    {
        minn = m;
        return;//不用往后搜索了 因为后面不如现在优 
    }
    for(i = 1;i <= 4;i ++)
    {
        int tx = x + xx[i];//横坐标更新
        int ty = y + yy[i];//纵坐标更新
        if(tx > 0 && tx <= 4 && ty > 0 && ty <= 4 && map[tx][ty] == b)//边界判断 
        {
            map[x][y] = map[tx][ty];
            map[tx][ty] = 0;//走完之后map[tx][ty]变空  map[x][y]变map[tx][ty]
            if(b == 1)
            b = 2;//黑白交替走 上次走黑 下次走白 
            else b = 1;
            for(j = 1;j <= 4;j ++)//因为空格又不止一个 所以 找空格 
            for(k = 1;k <= 4;k ++)
            if(!map[j][k])//空格的值为0 
                Dfs(j,k,num+1,b);
            map[tx][ty] = map[x][y];//回溯 
            map[x][y] = 0;
            if(b == 1)  b = 2;
            else    b = 1;
        }
    }
}
int main()
{
    int i,j;
    for(i = 1;i <= 4;i ++)
    for(j = 1;j <= 4;j ++)
    {
        cin >> s;
        if(s == ‘W‘) map[i][j] = 1;//白棋 
        if(s == ‘B‘) map[i][j] = 2;//黑棋 
    }
    for(i = 1;i <= 4;i ++)
    for(j = 1;j <= 4;j ++)
    if(!map[i][j])//空格 
    {
        Dfs(i,j,0,1);//先走白棋 
        Dfs(i,j,0,2);//先走黑棋 
    }
    cout<<minn<<endl;
    return 0;
}