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();
}