UVa - 439 - Knight Moves(bfs求最短路)

#include<iostream>
#include<queue>
#include<string>
#include<cstring>
using namespace std;
int x1,y1,x2,y2,ans;
string str1,str2;
//用来标记是否已经访问过
int vis[10][10];
//八个方向
int dis[8][2] = {-1,2 ,1,2 ,2,1 ,2,-1 ,1,-2 ,-1,-2 ,-2,-1 ,-2,1};
struct Step{
int x,y,steps; //x和y表示左边点，steps表示到达该点的最少步数
};

void bfs(){
//起点==终点
if(x1==x2&&y1==y2){
ans = 0;
return ;
}
//起点入队
queue<Step>q;
Step s1;
s1.x=x1;
s1.y=y1;
s1.steps=0;
vis[x1][y1] = 1;
q.push(s1);

while(!q.empty()){

Step s2;

//八个方向遍历
for(int i=0 ;i<8 ;i++){
s2.x = q.front().x+dis[i][0];
s2.y = q.front().y+dis[i][1];
s2.steps = q.front().steps+1;
if(vis[s2.x][s2.y]==1||s2.x<0||s2.y<0||s2.x>7||s2.y>7){
continue;
}
else{
vis[s2.x][s2.y]=1;
q.push(s2);
//如果遍历中到达了终点 则跳出
if(s2.x==x2&&s2.y==y2){
break;
}
}

}
//获取答案 跳出while循环
if(s2.x==x2&&s2.y==y2){
ans = s2.steps;
break;
}

q.pop();
}

}

int main(){
while(cin>>str1>>str2){
memset(vis,0,sizeof(vis));
x1 = (int)str1[0]-'a';
y1 = (int)str1[1]-'0'-1;
x2 = (int)str2[0]-'a';
y2 = (int)str2[1]-'0'-1;
bfs();
cout<<"To get from "<<str1<<" to "<<str2<<" takes "<<ans<<" knight moves.\n";
}

return 0;
}