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：

 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

原文地址：https://www.cnblogs.com/New-ljx/p/11334853.html