时间:2022-10-01 13:35:10 | 栏目:C代码 | 点击:次
马踏棋盘,用1枚马走遍棋盘。我用一个二维数组记录模拟的整个路径,x为列,y为行,以顺时针的方式寻找下一格,算法比较简单,就通过递归和循环回溯即可,就是如果是8*8的数组,最坏可能执行8^(x*y)次,耗时长到怀疑人生。
#include<iostream> #define X 5 #define Y 5 void ShowResult(); using namespace std; int chess[Y][X]={ 0 }; int counter=0; int Next(int* x,int* y,int where){ switch(where){ case 0: if(*x+1<X&&*y-2>=0&&chess[*y-2][*x+1]==0){ *x+=1; *y-=2; return 1; } break; case 1: if(*x+2<X&&*y-1>=0&&chess[*y-1][*x+2]==0){ *x+=2; *y-=1; return 1; } break; case 2: if(*x+2<X&&*y+1<Y&&chess[*y+1][*x+2]==0){ *x+=2; *y+=1; return 1; } break; case 3: if(*x+1<X&&*y+2<Y&&chess[*y+2][*x+1]==0){ *x+=1; *y+=2; return 1; } break; case 4: if(*x-1>=0&&*y+2<Y&&chess[*y+2][*x-1]==0){ *x-=1; *y+=2; return 1; } break; case 5: if(*x-2>=0&&*y+1<Y&&chess[*y+1][*x-2]==0){ *x-=2; *y+=1; return 1; } break; case 6: if(*x-2>=0&&*y-1>=0&&chess[*y-1][*x-2]==0){ *x-=2; *y-=1; return 1; } break; case 7: if(*x-1>=0&&*y-2>=0&&chess[*y-2][*x-1]==0){ *x-=1; *y-=2; return 1; } break; } return 0; } int Explore(int x,int y){ int x1=x; int y1=y; int flag; int where=0; counter++; chess[y][x]=counter; if(counter==X*Y){ return 1; } flag=Next(&x1,&y1,where); while(flag==0&&where<7){ where++; flag=Next(&x1,&y1,where); } while(flag){ if(Explore(x1,y1)==1){ return 1; } else{ x1=x; y1=y; where++; flag=Next(&x1,&y1,where); while(flag==0&&where<7){ where++; flag=Next(&x1,&y1,where); } } } if(flag==0){ chess[y][x]=0; counter--; } return 0; } void ShowResult(){ for(int i=0;i<Y;i++){ for(int j=0;j<X;j++){ cout.width(4); cout<<chess[i][j]<<' '; } cout<<endl; } cout<<endl; } int main(){ int start=clock(); int result=Explore(2,1); int end=clock(); if(result){ ShowResult(); } else{ cout<<"have no path!"<<endl; } cout<<"spend time:"<<(end-start)/CLOCKS_PER_SEC<<" s"<<endl; return 0; }