Eight HDU - 1043 八数码 A*算法

康托展开判重。也可以用打表和双向bfs做。
第一个A*搜索，A*是一种启发式搜索，g为已花代价，h为估计的剩余代价，而A*是根据f=g+h作为估价函数进行排列，也就是优先选择可能最优的节点进行扩展。可以对f进行关键字耗时600ms， 如果对h作为第一关键字， g作为第二关键字耗时170ms。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <queue>

using namespace std;

struct node
{
    int a[3][3];
    int c, x, y;
    int f, g, h;
    bool operator < (const node &F) const
    {
        if (h == F.h)
            return g > F.g;
        return h > F.h;
    }
}s;

int factor[] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880};
int dx[] = {0, 0, 1, 0, -1};
int dy[] = {0, 1, 0, -1, 0};
int f[363000];
int path[363000];

int cantor(int t[][3])
{
    int x[9];
    for (int i = 0; i < 9; i++)
        x[i] = t[i / 3][i % 3];
    int ans = 0;
    for (int i = 0; i < 9; i++)
    {
        int cnt = 0;
        for (int j = i + 1; j < 9; j++)
            if (x[i] > x[j])
                cnt++;
        ans += cnt * factor[9 - i - 1];
    }
    return ans;
}

int geth(int t[][3])
{
    int ans = 0;
    for (int i = 0; i < 3; i++)
        for (int j = 0; j < 3; j++)
        {
            if (!t[i][j]) continue;
            int x = (t[i][j] - 1) / 3, y = (t[i][j] - 1) % 3;
            ans += abs(x - i) + abs(y - j);
        }
    return ans;
}

bool judge(int t[][3])
{
    int x[9];
    for (int i = 0; i < 9; i++)
        x[i] = t[i / 3][i % 3];
    int cnt = 0;
    for (int i = 0; i < 9; i++)
        for (int j = i + 1; j < 9; j++)
            if (x[i] > x[j] && x[i] && x[j])
                cnt++;
    return cnt & 1;
}

void print(int x)
{
    if (s.c != x)
    {
        print(f[x]);
        if (path[x] == 1) cout << "r";
        if (path[x] == 2) cout << "d";
        if (path[x] == 3) cout << "l";
        if (path[x] == 4) cout << "u";
    }
}

void Astar()
{
    priority_queue<node> q;
    bool vis[363000] = {0};
    int t[][3] = {1, 2, 3, 4, 5, 6, 7, 8, 0};
    int target = cantor(t);
    s.c = cantor(s.a);
    vis[s.c] = true;
    q.push(s);
    while (!q.empty())
    {
        node cur = q.top(); q.pop();
        if (target == cur.c)
        {
            print(cur.c);
            cout << endl;
            break;
        }
        for (int i = 1; i <= 4; i++)
        {
            node c = cur; c.x += dx[i], c.y += dy[i];
            if (0 <= c.x && c.x < 3 && 0 <= c.y && c.y < 3)
            {
                swap(c.a[c.x][c.y], c.a[cur.x][cur.y]);
                c.c = cantor(c.a);
                if (!vis[c.c])
                {
                    vis[c.c] = true;
                    c.g++; c.h = geth(c.a); c.f = c.g + c.h;
                    f[c.c] = cur.c;
                    path[c.c] = i;
                    q.push(c);
                }
            }
        }
    }
}

int main()
{
    ios::sync_with_stdio(0);
    char ch;
    while (cin >> ch)
    {
        if (ch == 'x')
            s.x = 0, s.y = 0, s.a[0][0] = 0;
        else
            s.a[0][0] = ch - '0';
        for (int i = 1; i < 9; i++)
        {
            cin >> ch;
            if (ch == 'x')
                s.x = i / 3, s.y = i % 3, s.a[i / 3][i % 3] = 0;
            else
                s.a[i / 3][i % 3] = ch - '0';
        }
        s.g = 0, s.h = geth(s.a), s.f = s.g + s.h;
        if (judge(s.a))
            cout << "unsolvable" << endl;
        else
            Astar();
    }

    return 0;
}