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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 500000using 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();}

 

poj 1077 Eight(A*)