hdu 1372 Knight Moves

来源：互联网发布：复旦金融硕士知乎编辑：程序博客网时间：2024/05/21 10:45

广搜, 那个Knight是走"日"字的, 就像中古想起里面的马走法一样哦, 那么它在不出范围的情况下就有8个方向哦.

#include <stdio.h>#include <string.h>#include <queue>using namespace std;struct cell {    int x, y;    int step;};int dir[8][2] = {-2, -1, -2, 1, -1, 2, 1, 2, 2, 1, 2, -1, 1, -2, -1, -2}; // 8个方向哦bool visited[9][9];cell start, end, go, to;int bfs() {    int i;    queue<cell> Q;    Q.push(start);    memset(visited, false, sizeof(visited));    visited[start.x][start.y] = true;    while (!Q.empty()) {        go = Q.front();        Q.pop();        for (i = 0; i < 8; i++) {            to.x = go.x + dir[i][0];            to.y = go.y + dir[i][1];            to.step = go.step;            if (to.x == end.x && to.y == end.y) {                return to.step + 1;            }            if (to.x >= 1 && to.x <= 8 && to.y >= 1 && to.y <= 8 && !visited[to.x][to.y]) {                ++to.step;                visited[to.x][to.y] = true;                Q.push(to);            }        }    }    return 0;}int main() {    char s[3], e[3];    int ans;    while (scanf("%s %s", s, e) == 2) {        start.x = s[1] - '0';        start.y = s[0] - 'a' + 1;        start.step = 0;        end.x = e[1] - '0';        end.y = e[0] - 'a' + 1;        if (start.x == end.x && start.y == end.y) {            printf("To get from %s to %s takes 0 knight moves.\n", s, e);            continue ;        }        ans = bfs();        printf("To get from %s to %s takes %d knight moves.\n", s, e, ans);    }    return 0;}