四种寻路算法并比较 好久没搞这些东西了...想了十分钟才勉强回忆起来...写了三个钟头...好累啊...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[i]; //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[i]=abs(x+dx[i]-TargetX)+abs(y+dy[i]-TargetY); for (i=0;i<4;i++) if (dis[i]<t) { order[0]=i+1; t=dis[i]; } if (Act[Level]==0) return order[0]; order[1]=-1; for (i=0;i<4;i++) if ((dis[i]==t)&&(i!=(order[0]-1))) { order[1]=i+1; break; } if (order[1]!=-1) { for (i=0;i<4;i++) if (dis[i]!=t) { order[tt]=i+1; tt++; } } else { for (i=0;i<4;i++) if (dis[i]!=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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