八数码问题

利用hash来进行节点判重

typedef int State[9];
const int MAXSTATE = 1000000;
State st[MAXSTATE], goal;
int dist[MAXSTATE];
int vis[36288], fact[9];

const int dx[] = {-1,1,0,0};
const int dy[] = {0,0,-1,1};


const int MAXHASHSIZE = 1000003;
int head[MAXHASHSIZE], next[MAXHASHSIZE];

int hash(State& s) {
    int v = 0;
    for ( int i = 0; i <9; i++) v = v * 10 + s[i];
    return v % MAXHASHSIZE;
}

bool try_to_insert(int s) {
    int h = hash(st[s]);
    int u = head[h];
    while(u) {
        if(memcmp(st[u],st[s], sizeof(st[s])) == 0) return 0;
        u = next[u];
    }        
    next[s] = head[h];
    head[h] = s;
    return 1;
}

int bfs() {
    init_lookup_table();
    int front = 1, rear = 2;
    while(front < rear) {
        State& s = st[front];
        if(memcmp(goal, s, sizeof(s)) == 0) return front;
        int z;
        for ( z = 0; z < 9; z++ ) if (!s[z]) break;
        int x = z%3, y = z/3; 
        for ( int d = 0; d < 4; d++ ) {
            int newx = x + dx[d];
            int newy = y + dy[d];
            int newz = newy * 3 + newx;
            if ( newx >= 0 && newx < 3 && newy >= 0 && newy < 3) {
                State& t = st[rear];
                memcpy(&t, &s, sizeof(s));
                t[newz] = s[z];
                t[z] = s[newz];
                dist[rear] = dist[front] + 1;
                if(try_to_insert(rear)) rear++;
            }
        }
        front ++;
    }
    return 0;
}

int main() {
    for ( int i = 0; i < 9; i++) scanf("%d", &st[1][i]);
    for ( int i = 0; i < 9; i++) scanf("%d", &goal[i]);

    int ans = bfs();
    if (ans>0) printf("%d\n", dist[ans]);
    else printf("no\n");

    return 0;
}

下面是另一种节点查找表的实现, 把每个节点'变成'整数. 同时这也是一种完美的hash.

void init_lookup_table() {
    fact[0] = 1;
    for( int i = 1; i < 9; i++) fact[i] = fact[i-1]*i;
}

bool try_to_insert(int s) {
    int code = 0;
    for (int i = 0; i < 9; i++) {
        int cnt = 0;
        for ( int j = i + 1; j < 9; j++ ) if ( st[s][j] < st[s][i] ) cnt++;
        code += fact[8-i] * cnt;
    }
    if ( vis[code] ) return 0;
    return vis[code] = 1;
}