HDU 1043 eight (Astar 八数码)

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八数码问题是搜索算法中非常著名的问题，HDU这道八数码通过普通搜索无法通过，这里需要用到A* 算法或者双向广搜，这里我选择使用A* 算法来实现，借此机会学习一下A*算法
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<vector>#include<cmath>#include<map>#include<string>using namespace std;struct Node{    int maze[3][3];    int h, g;    int x, y;    int Hash;    bool operator<(const Node n1)const    {        return h != n1.h ? h>n1.h:g>n1.g;    }    bool check()    {        if (x >= 0 && x<3 && y >= 0 && y<3)            return true;        return false;    }} s, u, v, tt;int HASH[9] = { 1,1,2,6,24,120,720,5040,40320 };int destination = 322560;int vis[400000];int pre[400000];int way[4][2] = { { 0,1 },{ 0,-1 },{ 1,0 },{ -1,0 } };int get_hash(Node tmp){    int a[9], k = 0;    for (int i = 0; i<3; i++)        for (int j = 0; j<3; j++)            a[k++] = tmp.maze[i][j];    int res = 0;    for (int i = 0; i<9; i++)    {        int k = 0;        for (int j = 0; j<i; j++)            if (a[j]>a[i])                k++;        res += HASH[i] * k;    }    return res;}bool isok(Node tmp){    int a[9], k = 0;    for (int i = 0; i<3; i++)        for (int j = 0; j<3; j++)            a[k++] = tmp.maze[i][j];    int sum = 0;    for (int i = 0; i<9; i++)        for (int j = i + 1; j<9; j++)            if (a[j] && a[i] && a[i]>a[j])                sum++;    return !(sum & 1);}int get_h(Node tmp){    int ans = 0;    for (int i = 0; i<3; i++)        for (int j = 0; j<3; j++)            if (tmp.maze[i][j])                ans += abs(i - (tmp.maze[i][j] - 1) / 3) + abs(j - (tmp.maze[i][j] - 1) % 3);    return ans;}void astar(){    priority_queue<Node>que;    que.push(s);    while (!que.empty())    {        u = que.top();        que.pop();        for (int i = 0; i<4; i++)        {            v = u;            v.x += way[i][0];            v.y += way[i][1];            if (v.check())            {                swap(v.maze[v.x][v.y], v.maze[u.x][u.y]);                v.Hash = get_hash(v);                if (vis[v.Hash] == -1 && isok(v))                {                    vis[v.Hash] = i;                    v.g++;                    pre[v.Hash] = u.Hash;                    v.h = get_h(v);                    que.push(v);                }                if (v.Hash == destination)                    return;            }        }    }}void print(){    string ans;    ans.clear();    int nxt = destination;    while (pre[nxt] != -1)    {        switch (vis[nxt])        {        case 0:            ans += 'r';            break;        case 1:            ans += 'l';            break;        case 2:            ans += 'd';            break;        case 3:            ans += 'u';            break;        }        nxt = pre[nxt];    }    for (int i = ans.size() - 1; i >= 0; i--)        putchar(ans[i]);    puts("");}int main(){    std::ios::sync_with_stdio(false);    std::cin.tie(0);    string str;    while (getline(cin,str))    {        int k = 0;        memset(vis, -1, sizeof(vis));        memset(pre, -1, sizeof(pre));        for (int i = 0; i<3; i++)            for (int j = 0; j<3; j++)            {                if ((str[k] <= '9'&&str[k] >= '0') || str[k] == 'x')                {                    if (str[k] == 'x')                    {                        s.maze[i][j] = 0;                        s.x = i;                        s.y = j;                    }                    else                        s.maze[i][j] = str[k] - '0';                }                else                    j--;                k++;            }        if (!isok(s))        {            printf("unsolvable\n");            continue;        }        s.Hash = get_hash(s);        if (s.Hash == destination)        {            puts("");            continue;        }        vis[s.Hash] = -2;        s.g = 0;        s.h = get_h(s);        astar();        print();    }    return 0;}
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