HDU 1372 Knight Moves（bfs）

嗯...

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1372

这是一道很典型的bfs，跟马走日字一个道理，然后用dir数组确定骑士可以走的几个方向，然后从起点到终点跑一遍最典型的bfs即可...注意HDU的坑爹输入和输出...

AC代码：

 #include<cstdio>
 #include<iostream>
 #include<queue>
 #include<cstring>

 using namespace std;

 int ex, ey, sx, sy, step, g[][];

 struct node{
     int x, y, step;
 };

 int dir[][] = {{, }, {, }, {-, }, {, -}, {-, }, {, -}, {-, -}, {-, -}};
 //方向
 inline void bfs(){
     memset(g, , sizeof(g));
     queue<node> q;
     node now, next;
     now.x = sx;
     now.y = sy;
     now.step = ;
     g[now.x][now.y] = ;
     q.push(now);
     while(!q.empty()){
         now = q.front();
         q.pop();
         if(now.x == ex && now.y == ey){
             step = now.step;
             return;//找到终点
         }
         for(int i = ; i < ; i++){
             next.x = now.x + dir[i][];
             next.y = now.y + dir[i][];
             if(next.x >=  && next.x <=  && next.y >=  && next.y <=  && !g[next.x][next.y]){
                 next.step = now.step + ;
                 g[next.x][next.y] = ;
                 q.push(next);
             }
         }
     }
 }

 int main(){
     char c1, c2;
     int s, t;
     while(~scanf("%c%d %c%d", &c1, &s, &c2, &t)){
         getchar();
         sx = c1 - 'a' + ;
         sy = s;
         ex = c2 - 'a' + ;
         ey = t;//起终点
         bfs();
         printf("To get from %c%d to %c%d takes %d knight moves.\n", c1, s, c2, t, step);
     }
     return ;
 }

AC代码