八数码之A*解决方法

来源：互联网发布：就业前景好的专业知乎编辑：程序博客网时间：2024/05/28 16:06

学习过A*算法后，通过计算出每个节点的f值，在把队列里的节点通过对f的大小进行排列。
f=h+g
g是已经走过的步数，h为还需要最少的步数。

#include<iostream>using namespace std;const int MAXX = 1000000;struct Node{    char tile[10];    int pos;    int parents;    int g,f;    int pre;};int cmp(const void*a, const void *b){    if ((*(Node*)a).f == (*(Node*)b).f)        return ((*(Node*)a).g == (*(Node*)b).g);    return ((*(Node*)a).f - (*(Node*)b).f);}int _move[4][2] = { {0,1},{0,-1},{1,0},{-1,0} };char pre[4] = { 'r', 'l', 'd', 'u' };Node qu[MAXX];int front, tail,b;char _start[10],_end[10]="12345678x";void init(){    front = tail = 0;    qu[tail++].parents = -1;    strcpy(qu[0].tile, _start);    qu[0].pos = strchr(_start, 'x') - _start;    qu[0].g = 0;    b = 0;}int isexit(Node node){    for (int i = 0; i<tail; i++){        if (strcmp(qu[i].tile, node.tile) == 0) return 1;    }    return 0;}int getF(Node node){    int temp = 0;    for (int i = 0; i < 9; i++){        if (node.tile[i] != _end[i]) temp++;    }    return temp;}void show(Node temp){    if (temp.parents != 0){        show(qu[temp.parents]);    }    cout << pre[temp.pre];}void dfs(){    Node temp,now;    int x, y;    while (tail-front>0)    {        qsort(qu + front, tail - front, sizeof(Node), cmp);        now = qu[front++];        //找到答案        if (strcmp(now.tile, _end) == 0){            b = 1;            show(now);            cout << endl;            return;        }        temp.g = now.g + 1;        for (int i = 0; i < 4; i++){            strcpy(temp.tile, now.tile);            x = now.pos / 3 + _move[i][0];            y = now.pos % 3 + _move[i][1];            if (x < 0 || x >= 3 || y < 0 || y >= 3) continue;            temp.pos = x * 3 + y;            temp.tile[now.pos] = now.tile[temp.pos];            temp.tile[temp.pos] = 'x';            if (isexit(temp)){                continue;            }            temp.parents = front - 1;            temp.pre = i;            temp.f = temp.g+getF(temp);            qu[tail++] = temp;        }    }}int main(){    char str[50];    while (gets(str)){        int ans= 0;        for (int i = 0; str[i]; i++){            if (str[i] != ' ') _start[ans++] = str[i];        }        if (strcmp(_start, _end) == 0) cout << "lr" << endl;        else        {            init();            dfs();            if (!b) cout << "unsolvable" << endl;        }    }    return 0;}

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