A星算法，找寻最短路径

来源：互联网发布：linux运维工资高吗编辑：程序博客网时间：2024/06/05 15:17
#include <iostream>#include <cstdio>#include <queue>#include <algorithm>#include <cmath>#define N 1000#define inf 1<<30;using namespace std;/*a星算法，找寻最短路径算法核心：有两个表open表和close表将方块添加到open列表中，该列表有最小的和值。且将这个方块称为S吧。将S从open列表移除，然后添加S到closed列表中。对于与S相邻的每一块可通行的方块T：如果T在closed列表中：不管它。如果T不在open列表中：添加它然后计算出它的和值。如果T已经在open列表中：当我们使用当前生成的路径到达那里时，检查F（指的是和值）是否更小。如果是，更新它的和值和它的前继。F = G + H        (G指的是从起点到当前点的距离，而H指的是从当前点到目的点的距离（移动量估算值采用曼哈顿距离方法估算）*/int map[6][7];     //0表示是路，1表示有阻碍物int xstart, ystart, xend, yend;       //(x1,y1)起点，（x2, y2)目的点int close[6][7];  //0表示不在，1表示在int n;void astar(int , int );bool check(int x, int y);int panyical(int x, int y);void print2();struct point{int x, y;int f, g, h;int prex, prey;   //上一个点的x,ypoint(int x0, int y0, int g0, int h0){x = x0;y = y0;g = g0;h = h0;f = g + h;}point(){}}open[N];void print1(point);        //逆向反推void init(){xstart = 3, ystart = 1;   //起点xend = 4, yend = 5;   //目的点map[4][1] = map[1][3] = map[2][3] = map[3][3] = map[4][3] = 1;   //设置阻隔物astar(xstart,ystart);}point minfpoint(){int flag;int minf = inf;for(int t=0; t<n; ++t){if(close[open[t].x][open[t].y] == 0 && open[t].f <= minf){flag = t;minf = open[t].f;}}return open[flag];}void update(int x, int y, point &s){for(int t=0; t<n; ++t){if(open[t].x == x && open[t].y == y)         //如果该点已经在open表中存在的话，则更新值{int k = s.g+1+panyical(x, y);if(open[t].f > k){open[t].f = k;open[t].prex = s.x;open[t].prey = s.y;}return ;}}open[n] = point(x, y, s.g+1, panyical(x,y));open[n].prex = s.x;open[n].prey = s.y;n++;}void astar(int x, int y){point s = point(x, y, 0, 0);n = 0;while(1){close[s.x][s.y] = 1;if(s.x == xend && s.y == yend){break;}if(check(s.x+1, s.y)){update(s.x+1, s.y, s);}if(check(s.x-1, s.y)){update(s.x-1, s.y, s);}if(check(s.x, s.y+1)){update(s.x, s.y+1, s);}if(check(s.x, s.y-1)){update(s.x, s.y-1, s);}s = minfpoint();}printf("min:%d g:%d h:%d\n", s.f, s.g, s.h); //print1(s);print2();}void print1(point p){point s  = p;while(1){printf("(%d,%d  g:%d, h:%d f:%d)->", s.x, s.y, s.g, s.h, s.f);int prex = s.prex;int prey = s.prey;if(prex == xstart && prey == ystart){break;}for(int t=0; t<n; ++t){if(open[t].x == prex && open[t].y == prey){s = open[t];break;}}}}void print2(){for(int t=0; t<n; ++t){point s = open[t];printf("(%d,%d  g:%d, h:%d f:%d)\n", s.x, s.y, s.g, s.h, s.f);}}int panyical(int x, int y){return abs(xend-x) + abs(yend-y);}bool check(int x, int y){if(x<0 || x>5 || y<0 || y>6 || map[x][y]==1 || close[x][y]==1){return false;}return true;}int main(){init();return 0;}
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