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  • 八数码问题

    //八数码问题
    typedef int State[9];
    const int maxstate = 1000000;
    State st[maxstate], goal//所有转换的状态和终点状态
    int dist[maxstate];
    const int dx[] = {-1, 1, 0, 0};
    const int dy[] = {0, 0, -1, 1};
    
    int bfs() {
        init_lookup_table();
        int front = 1, rear = 2;
        while(front < rear) {
            State &s = st[front];//弹出队头状态
            if (memcmp(goal, s, sizeof(s)) == 0) return front;//找到目标状态
            int z;
            for (z = 0; z < 9; z ++) if(!s[z]) break;//找到0的位置;
            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 = newx * 3 + newy;
                if (newx >=0 && newx < 3 && newy >=0 && newy <3) {
                    steat &t = st[rear]; //拓展新节点
                    memcpy(&t, &s, sizeof[t]);
                    t[newz] = s[z];
                    t[z] = s[newz];
                    dist[rear] = dist[front] + 1;
                    if(try_to_instert(rear)) rear++;//如果可以插入节点,则将新节点拓展
                }
            }
            front ++;
        }
        return 0;
    }
    
    //用hashtable存储已遍历节点
    const int hashsize = 1000003;
    int head[hashsize], next[maxstate];
    void init_lookup_table() {
        memset(head, 0 ,sizeof(head));
    }
    int hash(State &s) {
        int v = 0;
        for (int i = 0; i < 9; i ++) {
            v =v * 10 + s[i];
        }
        return v % hashsize;
    }
    int try_to_instert(int s) {
        int h = hash(st[s]);
        int u = head[h];
        while(u) {
            if(memcmp(st[u], st[s], sizeof(st[u])) == 0) 
                return false;
            u = next[u];
        }
        next[s] = head[h];
        head[h] = s;
        return 1;
    }
    
    //将节点转换为9位十进制整数,并用set统计。
    set<int> vis;
    void init_lookup_table() {
        vis.clear();
    }
    int try_to_instert(int s) {
        int v = 0;
        for (int i = 0; i < 9; i++) {
            v = v * 10 +st[s][i];
        }
        if(vis.count(v)) return 0;
        vis.insert(v);
        return 1;
    }
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  • 原文地址:https://www.cnblogs.com/lichongjie/p/6863621.html
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