四种寻路算法计算步骤比较

来源：互联网发布：seo跟sem的区别编辑：程序博客网时间：2024/05/19 09:37

四种算法是DFS,BFS,Heuristic DFS, Heuristic BFS (A*)

用了两张障碍表，一张是典型的迷宫：

char Block[SY][SX]=

{{1,1,1,1,1,1,1,1,1,1,1 },

{1,0,1,0,1,0,0,0,0,0,1 },

{1,0,1,0,0,0,1,0,1,1,1 },

{1,0,0,0,1,0,1,0,0,0,1 },

{1,0,1,1,0,0,1,0,0,1,1 },

{1,0,1,0,1,1,0,1,0,0,1 },

{1,0,0,0,0,0,0,0,1,0,1 },

{1,0,1,0,1,0,1,0,1,0,1 },

{1,0,0,1,0,0,1,0,1,0,1 },

{1,1,1,1,1,1,1,1,1,1,1 }};

第二张是删掉一些障碍后的:

char Block[SY][SX]=

{{1,1,1,1,1,1,1,1,1,1,1 },

{1,0,1,0,1,0,0,0,0,0,1 },

{1,0,1,0,0,0,1,0,1,1,1 },

{1,0,0,0,0,0,1,0,0,0,1 },

{1,0,0,1,0,0,1,0,0,1,1 },

{1,0,1,0,0,1,0,1,0,0,1 },

{1,0,0,0,0,0,0,0,1,0,1 },

{1,0,1,0,0,0,1,0,1,0,1 },

{1,0,0,1,0,0,1,0,0,0,1 },

{1,1,1,1,1,1,1,1,1,1,1 }};

结果：

尝试节点数合法节点数步数

深度优先 416/133 110/43 19/25

广度优先 190/188 48/49 19/15

深度+启发 283/39 82/22 19/19

广度+启发 189/185 48/49 19/15

所以可以看出深度+启发是最好的，效率高路径也挺短。A*第一是不真实二是慢三是空间消耗较大。

附：dfs+heu的源程序，bc++ 3.1通过

#include <iostream.h>
#include <memory.h>
#include <stdlib.h>
#define SX 11 //宽
#define SY 10 //长
int dx[4]={0,0,-1,1}; //四种移动方向对x和y坐标的影响
int dy[4]={-1,1,0,0};
/*char Block[SY][SX]= //障碍表
{{ 1,1,1,1,1,1,1,1,1,1,1 },
{ 1,0,1,0,1,0,0,0,0,0,1 },
{ 1,0,1,0,0,0,1,0,1,1,1 },
{ 1,0,0,0,0,0,1,0,0,0,1 },
{ 1,0,0,1,0,0,1,0,0,1,1 },
{ 1,0,1,0,0,1,0,1,0,0,1 },
{ 1,0,0,0,0,0,0,0,1,0,1 },
{ 1,0,1,0,0,0,1,0,1,0,1 },
{ 1,0,0,1,0,0,1,0,0,0,1 },
{ 1,1,1,1,1,1,1,1,1,1,1 }};*/
char Block[SY][SX]= //障碍表
{{ 1,1,1,1,1,1,1,1,1,1,1 },
{ 1,0,1,0,1,0,0,0,0,0,1 },
{ 1,0,1,0,0,0,1,0,1,1,1 },
{ 1,0,0,0,1,0,1,0,0,0,1 },
{ 1,0,1,1,0,0,1,0,0,1,1 },
{ 1,0,1,0,1,1,0,1,0,0,1 },
{ 1,0,0,0,0,0,0,0,1,0,1 },
{ 1,0,1,0,1,0,1,0,1,0,1 },
{ 1,0,0,1,0,0,1,0,1,0,1 },
{ 1,1,1,1,1,1,1,1,1,1,1 }};
int MaxAct=4; //移动方向总数
char Table[SY][SX]; //已到过标记
int Level=-1; //第几步
int LevelComplete=0; //这一步的搜索是否完成
int AllComplete=0; //全部搜索是否完成
char Act[1000]; //每一步的移动方向，搜索1000步，够了吧？
int x=1,y=1; //现在的x和y坐标
int TargetX=9,TargetY=8; //目标x和y坐标
int sum1=0,sum2=0;
void Test( );
void Back( );
int ActOK( );
int GetNextAct( );
void main( )
{
memset(Act,0,sizeof(Act)); //清零
memset(Table,0,sizeof(Table));
Table[y][x]=1; //做已到过标记
while (!AllComplete) //是否全部搜索完
{
Level++;LevelComplete=0; //搜索下一步
while (!LevelComplete)
{
Act[Level]=GetNextAct( ); //改变移动方向
if (Act[Level]<=MaxAct)
sum1++;
if (ActOK( )) //移动方向是否合理
{
sum2++;
Test( ); //测试是否已到目标
LevelComplete=1; //该步搜索完成
}
else
{
if (Act[Level]>MaxAct) //已搜索完所有方向
Back( ); //回上一步
if (Level<0) //全部搜索完仍无结果
LevelComplete=AllComplete=1; //退出
}
}
}
}
void Test( )
{
if ((x==TargetX)&&(y==TargetY)) //已到目标
{
for (int i=0;i<=Level;i++)
cout<<(int)Act; //输出结果
cout<<endl;
cout<<Level+1<<" "<<sum1<<" "<<sum2<<endl;
LevelComplete=AllComplete=1; //完成搜索
}
}
int ActOK( )
{
int tx=x+dx[Act[Level]-1]; //将到点的x坐标
int ty=y+dy[Act[Level]-1]; //将到点的y坐标
if (Act[Level]>MaxAct) //方向错误？
return 0;
if ((tx>=SX)||(tx<0)) //x坐标出界？
return 0;
if ((ty>=SY)||(ty<0)) //y坐标出界？
return 0;
if (Table[ty][tx]==1) //已到过？
return 0;
if (Block[ty][tx]==1) //有障碍？
return 0;
x=tx;
y=ty; //移动
Table[y][x]=1; //做已到过标记
return 1;
}
void Back( )
{
x-=dx[Act[Level-1]-1];
y-=dy[Act[Level-1]-1]; //退回原来的点
Table[y][x]=0; //清除已到过标记
Act[Level]=0; //清除方向
Level--; //回上一层
}
int GetNextAct( ) //找到下一个移动方向。这一段程序有些乱，
//仔细看!
{
int dis[4];
int order[4];
int t=32767;
int tt=2;
for (int i=0;i<4;i++)
dis=abs(x+dx-TargetX)+abs(y+dy-TargetY);
for (i=0;i<4;i++)
if (dis<t)
{
order[0]=i+1;
t=dis;
}
if (Act[Level]==0)
return order[0];
order[1]=-1;
for (i=0;i<4;i++)
if ((dis==t)&&(i!=(order[0]-1)))
{
order[1]=i+1;
break;
}
if (order[1]!=-1)
{
for (i=0;i<4;i++)
if (dis!=t)
{
order[tt]=i+1;
tt++;
}
}
else
{
for (i=0;i<4;i++)
if (dis!=t)
{
order[tt-1]=i+1;
tt++;
}
}
if (Act[Level]==order[0])
return order[1];
if (Act[Level]==order[1])
return order[2];
if (Act[Level]==order[2])
return order[3];
if (Act[Level]==order[3])
return 5;
}

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