POJ 2243 ZOJ 1091 UVaOJ 439 Knight moves(BFS广度优先搜索)

//典型的BFS（广度优先搜索）//马走日所需最少步数问题#include<iostream>using namespace std;int map[8][8];//记录步数数组int vis[8][8];//访问过的记录数组char a[3],b[3];int x1,y1,x2,y2;void set(int x1,int y1){//剪去不合理范围if(x1+1 <= 7 && y1+2 <= 7 && !vis[x1+1][y1+2])map[x1+1][y1+2] = map[x1][y1]+1;if(x1+1 <= 7 && y1-2 >= 0 && !vis[x1+1][y1-2])map[x1+1][y1-2] = map[x1][y1]+1;if(x1-1 >= 0 && y1+2 <= 7 && !vis[x1-1][y1+2])map[x1-1][y1+2] = map[x1][y1]+1;if(x1-1 >= 0 && y1-2 >= 0 && !vis[x1-1][y1-2])map[x1-1][y1-2] = map[x1][y1]+1;if(x1+2 <= 7 && y1+1 <= 7 && !vis[x1+2][y1+1])map[x1+2][y1+1] = map[x1][y1]+1;if(x1-2 >= 0 && y1+1 <= 7 && !vis[x1-2][y1+1])map[x1-2][y1+1] = map[x1][y1]+1;if(x1+2 <= 7 && y1-1 >= 0 && !vis[x1+2][y1-1])map[x1+2][y1-1] = map[x1][y1]+1;if(x1-2 >= 0 && y1-1 >= 0 && !vis[x1-2][y1-1])map[x1-2][y1-1] = map[x1][y1]+1;}void search(int x1,int y1,int x2,int y2){int step = 1;vis[x1][y1] = 1;set(x1,y1);//广度优先搜索过程while(map[x2][y2] == 0){for(int i = 0;i < 8;++i)for(int j = 0;j < 8;++j)if(map[i][j] == step){vis[i][j] = 1;set(i,j);}++step;}}int main(){while(cin >> a >> b){memset(map,0,sizeof(map));//初始化memset(vis,0,sizeof(vis));//初始化x1 = a[0] - 'a';y1 = a[1] - '1';x2 = b[0] - 'a';y2 = b[1] - '1';if(strcmp(a,b) == 0)cout << "To get from "<< a <<" to "<< b << " takes 0 knight moves." <<endl;else{search(x1,y1,x2,y2);cout << "To get from "<< a <<" to "<< b << " takes "<< map[x2][y2] <<" knight moves." << endl;}}return 0;}