UVa439 Knight Moves (BFS求最短路)

Posted 2020-08-06

链接：http://acm.hust.edu.cn/vjudge/problem/19436
分析：BFS跑一次最短路，状态转移有8个。

 1 #include <cstdio>
 2 #include <queue>
 3 #include <cstring>
 4 using namespace std;
 5 
 6 struct Point {
 7     char r, c;
 8     Point(char r = ‘ ‘, char c = ‘ ‘): r(r), c(c) {};
 9 };
10 
11 int vis[60][110], d[60][110];
12 
13 const int dir[8][2] = {
14                     {2, -1}, {1, -2},
15                     {-1, -2}, {-2, -1},
16                     {-2, 1}, {-1, 2},
17                     {1, 2}, {2, 1} };
18 int main() {
19     char s[3], t[3];
20     while (scanf("%s%s", s, t) == 2) {
21         memset(vis, 0, sizeof(vis));
22         Point p; p.r = s[1]; p.c = s[0];
23         d[p.r][p.c] = 0;
24         queue<Point> q;
25         q.push(p);
26         while (!q.empty()) {
27             Point u = q.front(); q.pop();
28             if (u.r == t[1] && u.c == t[0]) break;
29             if (vis[u.r][u.c]) continue; else vis[u.r][u.c] = 1;
30             for (int i = 0; i < 8; i++) {
31                 char nr = u.r + dir[i][0], nc = u.c + dir[i][1];
32                 if (nr >= ‘1‘ && nr <= ‘8‘ && nc >= ‘a‘ && nc <= ‘h‘) {
33                     d[nr][nc] = d[u.r][u.c] + 1;
34                     q.push(Point(nr, nc));
35                 }
36             }
37         }
38         printf("To get from %s to %s takes %d knight moves.\n", s, t, d[t[1]][t[0]]);
39     }
40     return 0;
41 }