UVA 439-Knight Moves

题目大意：给出国际象棋棋盘俩个坐标，求马从坐标1到坐标2的最短步数

解题思路：bfs寻找最短路

#include <stdio.h>#include <iostream>#include <queue>#include <string.h>using namespace std;struct kni{    int x, y, d;};int dir[8][2] = {{1, 2}, {-1, 2}, {2, 1}, {2, -1}, {-2, 1}, {-2, -1}, {1, -2}, {-1, -2}};queue<kni> que;int map[9][9];int bfs(int a, int b) {    while(!que.empty()) {        if(que.front().x == a && que.front().y == b)            return 0;        for(int i = 0; i < 8; i++) {            int c = que.front().x + dir[i][0], d = que.front().y + dir[i][1];            if(c > 0 && c <= 8 && d > 0 && d <= 8 && map[c][d] == 0) {                if(c == a && d == b)                    return que.front().d + 1;                else {                    map[que.front().x][que.front().y] = 1;                    kni kk;                    kk.x = c;                    kk.y = d;                    kk.d = que.front().d+1;                    que.push(kk);                }            }        }        que.pop();    }    return -1;}int main() {    char ch1, ch2;    int x1, x2, y1, y2;    while(cin >> ch1 >> y1 >> ch2 >> y2) {        while(!que.empty())            que.pop();        memset(map, 0, sizeof(map));        x1 = ch1 - 'a' + 1;        x2 = ch2 - 'a' + 1;        kni k;        k.x = x1;        k.y = y1;        k.d = 0;        que.push(k);        int re = bfs(x2, y2);        printf("To get from %c%d to %c%d takes %d knight moves.\n", ch1, y1, ch2, y2, re);    }    return 0;}