p-7-15最近距离

1. 题目描述

在一个游戏中，玩家处于一个如下所示12行12列的迷宫：

0,1,0,0,0,1,1,1,0,1,0,1

0,0,0,1,0,0,0,0,1,0,0,1

0,1,0,1,0,1,1,1,0,1,0,0

0,1,0,0,0,0,0,1,0,0,1,1

0,0,0,0,1,0,0,0,0,0,0,0

0,0,1,0,0,0,1,0,0,0,1,0

0,0,1,0,0,0,0,0,1,0,0,0

1,0,0,1,0,1,0,0,0,1,0,1

0,0,1,0,1,0,1,0,1,0,0,0

0,0,0,0,0,1,0,0,0,1,1,0

0,0,0,0,0,1,0,0,0,0,0,0

0,1,0,1,0,0,0,1,0,1,0,0

其中迷宫由0,1组成，0表示道路，1表示障碍物。

现在要根据玩家和游戏中被攻击的虚拟boss所在位置，给玩家以最近距离的提示。

最近距离：即玩家走到boss所走的最少步数。（注：路线中的一步是指从一个坐标点走到其上下左右相邻坐标点。）

2.输入格式:

输入4个整数a,b,c,d（即玩家和虚拟boss在迷宫中的坐标位置分别为(a,b) 、(c,d)）， 其中 0<=a,b,c,d<12。

3.输出格式:

输出在迷宫中从(a,b)出发到达(c,d)的最少步数，如果(a,b)永远无法到达(c,d)则输出10000。

4. 输入样例:

在这里给出一组输入。例如：

0 0 11 11

5. 输出样例:

在这里给出相应的输出。例如：

22

#include <bits/stdc++.h>
using namespace std;

int mp[12][12] =  { 0,1,0,0,0,1,1,1,0,1,0,1,

              0,0,0,1,0,0,0,0,1,0,0,1,

              0,1,0,1,0,1,1,1,0,1,0,0,

              0,1,0,0,0,0,0,1,0,0,1,1,

              0,0,0,0,1,0,0,0,0,0,0,0,

              0,0,1,0,0,0,1,0,0,0,1,0,

              0,0,1,0,0,0,0,0,1,0,0,0,

              1,0,0,1,0,1,0,0,0,1,0,1,

              0,0,1,0,1,0,1,0,1,0,0,0,

              0,0,0,0,0,1,0,0,0,1,1,0,

              0,0,0,0,0,1,0,0,0,0,0,0,

              0,1,0,1,0,0,0,1,0,1,0,0
            };//初始地图

int dist = 0, min_dist = 10000;
int a, b, c, d;
// (a, b)可以走的方向
int A[4] = {-1, 1, 0, 0};
int B[4] = {0, 0, -1, 1};

void dfs(int a, int b) {

  if(a == c && b == d) {
    if(dist < min_dist)
      min_dist = dist;
  } else {
    for (int i = 0; i < 4; i++) {
      if(a + A[i] >= 0 && a + A[i] < 12 && b + B[i] >= 0 && b + B[i] < 12 && !mp[a + A[i]][b + B[i]] && dist < min_dist) {
        mp[a + A[i]][b + B[i]] = 1; // 该位置由0置为1
        dist++; // 距离增加
        dfs( a + A[i], b + B[i]);
        dist--; // 该坐标的其它方向不通  要回溯，并且将该位置置为初始状态 
        mp[a + A[i]][b + B[i]] = 0;
      }
    }
  }
}

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0); // 优化代码
  
  cin >> a >> b >> c >> d;
  dfs( a, b);
  cout << min_dist;
  return 0;
}