uva 439 - Knight Moves-优快云博客

本文介绍了一种使用广度优先搜索（BFS）算法来解决国际象棋中骑士从一个位置走到另一个位置最少步数的问题。通过定义骑士的移动方式，并采用结构体存储坐标与步数，实现了对所有可能路径的有效探索。

2019独角兽企业重金招聘Python工程师标准>>>

先研究下国际象棋中马是怎么走的，然后bfs就哦了。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

int map[10][10];

typedef struct Point {
	int x;
	int y;
	int dis;
} Point;

typedef Point* Ptr;

Point queue[5000];

int dir[8][2] = { { -2, 1 }, { -2, -1 }, { -1, 2 }, { 1, 2 }, { -1, -2 }, { 1,
		-2 }, { 2, 1 }, { 2, -1 } };

int bfs(Point from, Point to) {
	if (from.x == to.x && from.y == to.y)
		return 0;

	int front = 0;
	int rear = 1;
	queue[0].x = from.x;
	queue[0].y = from.y;
	queue[0].dis = from.dis;
	while (front < rear) {
		Ptr p = &queue[front++];
		map[p->x][p->y] = 1;
		if (p->x == to.x && p->y == to.y)
			return p->dis;
		int k;
		int newX, newY;
		for (k = 0; k < 8; k++) {
			newX = p->x + dir[k][0];
			newY = p->y + dir[k][1];
			if (newX >= 1 && newX <= 8 && newY >= 1 && newY <= 8
					&& !map[newX][newY]) {
				queue[rear].x = newX;
				queue[rear].y = newY;
				queue[rear].dis = p->dis + 1;
				rear++;
			}
		}
	}
	return -1;
}

int main() {
	char start[5];
	char end[5];
	while (scanf("%s%s", start, end) != EOF) {
		memset(map, 0, sizeof(map));
		Point from, to;
		from.x = start[0] - 'a' + 1;
		from.y = start[1] - '0';
		from.dis = 0;
		to.x = end[0] - 'a' + 1;
		to.y = end[1] - '0';
		to.dis = -1;
		printf("To get from %s to %s takes %d knight moves.\n", start, end,
				bfs(from, to));
	}
	return 0;
}

转载于:https://my.oschina.net/jdflyfly/blog/283606