标签：break clu bre close input determine struct case sample
e2 e4 a1 b2 b2 c3 a1 h8 a1 h7 h8 a1 b1 c3 f6 f6Sample Output
To get from e2 to e4 takes 2 knight moves. To get from a1 to b2 takes 4 knight moves. To get from b2 to c3 takes 2 knight moves. To get from a1 to h8 takes 6 knight moves. To get from a1 to h7 takes 5 knight moves. To get from h8 to a1 takes 6 knight moves. To get from b1 to c3 takes 1 knight moves. To get from f6 to f6 takes 0 knight moves.
分析：简单的BFS求最短路，需要注意的是国际象棋中骑士和中国象棋中的马一样走日字
代码：
#include<iostream> #include<queue> #include<cstdio> using namespace std; const int N = 10; typedef struct { int x; int y; } P; int dx[] = {-1, 1, -2, 2, -2, 2, -1, 1}; int dy[] = {-2, -2, -1, -1, 1, 1, 2, 2}; int vis[N][N]; int d[N][N]; char s[5]; char e[5]; int sx, sy, ex, ey; int check(int x, int y) { return x >= 0 && x < 8 && y >= 0 && y < 8; } void bfs() { queue<P> que; P p; p.x = sx; p.y = sy; que.push(p); vis[sx][sy] = 1; while(que.size()) { P p = que.front(); que.pop(); int x = p.x; int y = p.y; if(x == ex && y == ey) { printf("To get from %s to %s takes %d knight moves.\n", s, e, d[x][y]); break; } for(int i = 0; i < 8; i++) { int nx = x + dx[i], ny = y + dy[i]; if(check(nx, ny) && !vis[nx][ny]) { vis[nx][ny] = 1; d[nx][ny] = d[x][y] + 1; P p; p.x = nx; p.y = ny; que.push(p); } } } } int main() { while(cin >> s >> e) { for(int i = 0; i < N; i++) { for(int j = 0; j < N; j++) d[i][j] = vis[i][j] = 0; } sx = s[1] - ‘0‘ - 1; sy = s[0] - ‘a‘; ex = e[1] - ‘0‘ - 1; ey = e[0] - ‘a‘; bfs(); } return 0; }
标签：break clu bre close input determine struct case sample
原文地址：https://www.cnblogs.com/kindleheart/p/9297373.html
