poj 1077 Eight（A*）

     经典的八数码问题，用来练习各种搜索=_=。这题我用的A*做的，A*的主要思想就是在广搜的时候加了一个估价函数，用来评估此状态距离最终状态的大概距离。这样就可以省下很多状态不用搜索。对于每个状态设置一个函数 h(x)，这就是估价函数了（可能名词不太对请见谅），再设置一个函数 g(x)， 这存的是初始状态到当前状态所用步数（或距离，视估价函数而定），再设函数 f(x) = g(x) + h(x)，我们对每个状态的 f(x)的值进行排序， 然后从当前 f(x) 最小的状态继续搜索，直到搜索到最终状态或者没有后继状态。

    上代码：

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <iostream>

#include <algorithm>

#include <cmath>

#include <queue>

#define N 10

#define M 500000

using namespace std;



struct sss

{

    int state[N];

    int num, place;

};

priority_queue<sss> q;



int f[M], g[M] = {0}, fa[M] = {0}, move[M] = {0};

sss begin, end;

int step[4][2] = {{-1,0},{0,-1},{0,1},{1,0}};

int color[M] = {0};



bool operator < (sss a, sss b) { return f[a.num] > f[b.num]; }

bool operator == (sss a, sss b) { return a.num == b.num; }



int h(sss a) // h(v) 

{

    int z = 0;

    for (int i = 1; i <= 3; ++i)

        for (int j = 1; j <= 3; ++j)

        {

            int x = (i-1)*3+j;

            z += abs(i-(a.state[x]-1)/3) + abs(j-(a.state[x]-(a.state[x]-1)/3*3));

        }

    return z;

}



int jiecheng[N] = {0,1,2,6,24,120,720,5040,40320,362880};

int make_num(sss a) // 康拓展开 

{

    int ans = 0; bool xiao[N] = {0};

    for (int i = 1; i <= 9; ++i)

    {

        int count = 0; xiao[a.state[i]] = 1;

        for (int j = 1; j < a.state[i]; ++j)

            if (!xiao[j]) count++;

        ans += count * jiecheng[N-i-1];

    }

    return ans;

}



bool in(sss now, int mo) // 移动可行 

{

    int x = (now.place-1)/3+1, y = now.place-(x-1)*3;

    x += step[mo-1][0]; y += step[mo-1][1];

    if (x < 1 || x > 3 || y < 1 || y > 3) return false;

    else return true;

}



void A_star() // A* 

{

    q.push(begin); f[begin.num] = h(begin);

    while (!q.empty())

    {

        sss u = q.top(); q.pop();

        if (u == end)

            return;

        for (int i = -3; i <= 3; i+=2)

        {

            if (!in(u,(i+5)/2)) continue;

            sss v; v = u; swap(v.state[u.place+i], v.state[u.place]);

            v.place = u.place + i; v.num = make_num(v);

            if (color[v.num] == 1)

            {

                if (g[u.num] + 1 < g[v.num])

                {

                    f[v.num] = f[v.num] - g[v.num] + g[u.num] + 1;

                    g[v.num] = g[u.num] + 1; move[v.num] = i;

                    q.push(v); fa[v.num] = u.num;

                }

            }

            else if (color[v.num] == 2)

            {

                if (g[u.num] + 1 < g[v.num])

                {

                    f[v.num] = f[v.num] - g[v.num] + g[u.num] + 1;

                    g[v.num] = g[u.num] + 1; fa[v.num] = u.num;

                    q.push(v); color[v.num] = 1; move[v.num] = i;

                }

            }

            else

            {

                g[v.num] = g[u.num] + 1;

                f[v.num] = h(v) + g[v.num];

                color[v.num] = 1; fa[v.num] = u.num;

                q.push(v); move[v.num] = i;

            }

        }

        color[u.num] = 2;

    }

}



char change(int x)

{

    if (x == -1) return 'l';

    else if (x == -3) return 'u';

    else if (x == 1) return 'r';

    else return 'd';

}



void print() // 输出 

{

    char s[M], snum = 0;

    int k = fa[end.num];

    s[++snum] = change(move[end.num]);

    while (k != begin.num)

    {

        s[++snum] = change(move[k]);

        k = fa[k];

    }

    for (int i = snum; i > 0; --i)

        printf("%c", s[i]);

    printf("\n");

}



int main()

{

    char s[2];

    for (int i = 1; i <= 9; ++i)

    {

        scanf("%s", s);

        if (s[0] != 'x') begin.state[i] = s[0] - '0';

        else

        {

            begin.state[i] = 9;

            begin.place = i;

        }

        end.state[i] = i;

    }

    begin.num = make_num(begin);

    end.num = make_num(end);

    A_star();

    print();

}