牛客假日团队赛6 F:Mud Puddles

链接：https://ac.nowcoder.com/acm/contest/993/F
来源：牛客网
 

题目描述

Farmer John is leaving his house promptly at 6 AM for his daily milking of Bessie. However, the previous evening saw a heavy rain, and the fields are quite muddy. FJ starts at the point (0, 0) in the coordinate plane and heads toward Bessie who is located at (X, Y) (-500 ≤ X ≤ 500; -500 ≤ Y ≤ 500). He can see all N (1 ≤ N ≤ 10,000) puddles of mud, located at points (Ai, Bi) (-500 ≤ Ai ≤ 500; -500 ≤ Bi ≤ 500) on the field. Each puddle occupies only the point it is on.
Having just bought new boots, Farmer John absolutely does not want to dirty them by stepping in a puddle, but he also wants to get to Bessie as quickly as possible. He's already late because he had to count all the puddles. If Farmer John can only travel parallel to the axes and turn at points with integer coordinates, what is the shortest distance he must travel to reach Bessie and keep his boots clean? There will always be a path without mud that Farmer John can take to reach Bessie.

输入描述:

* Line 1: Three space-separate integers: X, Y, and N.
* Lines 2..N+1: Line i+1 contains two space-separated integers: Ai and Bi

输出描述:

* Line 1: The minimum distance that Farmer John has to travel to reach Bessie without stepping in mud.

示例1

输入

复制

1 2 7
0 2
-1 3
3 1
1 1
4 2
-1 1
2 2

输出

复制

11

说明

Bessie is at (1, 2). Farmer John sees 7 mud puddles, located at (0, 2); (-1, 3); (3, 1); (1, 1); (4, 2); (-1, 1) and (2, 2).
The best path for Farmer John is (0, 0); (-1, 0); (-2, 0); (-2, 1); (-2, 2); (-2, 3); (-2, 4); (-1, 4); (0, 4); (0, 3); (1, 3); and (1, 2), finally reaching Bessie.

题意分析：

给出n个水坑的位置，当前人在0 0上，求到x y最少需要多少步

解题思路：

因为给出的坐标可能为负，所以把坐标都加上500后进行bfs。

#include <stdio.h>
#include <algorithm>
#include <queue>
using namespace std;
#define N 1020
int map[N][N], book[N][N];

struct dat {
	int x;
	int y;
	int temp;
}e;
queue<dat>q;
int main()
{
	int nexts[4][2] = { 1,0, -1,0, 0,1, 0,-1 };
	int n, m, i, j, a, b, x, y, tx, ty;
	scanf("%d%d%d", &x, &y, &n);
	x += 500;
	y += 500;
	for (i = 0; i < n; i++)
	{
		scanf("%d%d", &a, &b);
		map[a+500][b+500] = 1;
	}

	q.push(dat{ 500, 500, 0 });
	book[500][500] = 1;
	while (!q.empty())
	{
		e = q.front();
		q.pop();
		if (e.x == x && e.y == y)
			break;
		for (i = 0; i < 4; i++)
		{
			tx = e.x;
			ty = e.y;
			tx += nexts[i][0];
			ty += nexts[i][1];
			if (book[tx][ty] == 0 && map[tx][ty] == 0)
			{
				book[tx][ty] = 1;
				q.push(dat{ tx, ty, e.temp + 1 });
			}
		}
	}
	printf("%d
", e.temp);
	return 0;
}