HDU 1372 Knight Moves（bfs）

Posted 2022-06-30 new-ljx

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：

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<queue>
 4 #include<cstring>
 5 
 6 using namespace std;
 7 
 8 int ex, ey, sx, sy, step, g[10][10];
 9 
10 struct node
11     int x, y, step;
12 ;
13 
14 int dir[8][2] = 2, 1, 1, 2, -1, 2, 2, -1, -2, 1, 1, -2, -1, -2, -2, -1;
15 //方向 
16 inline void bfs()
17     memset(g, 0, sizeof(g));
18     queue<node> q;
19     node now, next;
20     now.x = sx;
21     now.y = sy;
22     now.step = 0;
23     g[now.x][now.y] = 1;
24     q.push(now);
25     while(!q.empty())
26         now = q.front();
27         q.pop();
28         if(now.x == ex && now.y == ey)
29             step = now.step;
30             return;//找到终点 
31         
32         for(int i = 0; i < 8; i++)
33             next.x = now.x + dir[i][0];
34             next.y = now.y + dir[i][1];
35             if(next.x >= 1 && next.x <= 8 && next.y >= 1 && next.y <= 8 && !g[next.x][next.y])
36                 next.step = now.step + 1;
37                 g[next.x][next.y] = 1;
38                 q.push(next);
39             
40         
41     
42 
43 
44 
45 int main()
46     char c1, c2;
47     int s, t;
48     while(~scanf("%c%d %c%d", &c1, &s, &c2, &t))
49         getchar();
50         sx = c1 - ‘a‘ + 1;
51         sy = s;
52         ex = c2 - ‘a‘ + 1;
53         ey = t;//起终点 
54         bfs();
55         printf("To get from %c%d to %c%d takes %d knight moves.\n", c1, s, c2, t, step);
56     
57     return 0;
58 

AC代码