maze

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Note: If we reach a grid where a trap is located, the trap will be triggered automatically and disappear immediately after triggering.

If you go to the i-th organ and step on any of the organs after i, you will fail.

Born in any institution except the first institution, fail 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
 
using namespace std;
 
const int N = 110;
 
char a[N][N]; // store the map
int vis[N][N]; // vis[i][j] = 1 if the point has been traversed, 0 has not been traversed
int mis[N][N]; // mis[i][j] = 1 This point will trigger the trap in the future, so it cannot be traversed now  
 
typedef struct point {
   int x, y, step;
}P;
 
int n, m, k;

// 8 directions
int road[8][2] = {-1,-1, -1,0, -1,1,
                  0,-1, 0, 1,
                  1,-1, 1,0, 1,1};
 
queue <P> Q;
P score[N];
 
// Determine whether the item is outside the picture
bool judge(P tem) {
   if(tem.x < 0 || tem.x >= n || tem.y < 0 || tem.y >= m)
       return false;
   return  true;
}
 
// Determine whether it is possible to walk diagonally to the tem
bool wtwalk(int i, P tem) {
   P tem1, tem2;
   has1 = has;
   has2 = has;
 
   if(i == 0) {
      tem1.y ++;
      tem2.x++;
   }
   else if(i == 2) {
      item1.y --;
      tem2.x++;
   }
   else if(i == 5) {
      tem1.x -;
      tem2.y ++;
   }
   else {
      tem1.x--;
      item2.y--;
   }
 
   if(a[tem1.x][tem1.y] == '#' && a[tem2.x][tem2.y] == '#')
      return false;
   return true;
}

// bfs Sta is the starting point, End is the end point
// If the end point can be reached, return the number of steps
// if not return -1
int bfs(P Sta, P End) {
    memset(vis, 0, sizeof(vis));
    vis [Sta.x] [Sta.y] = 1;
    while(!Q.empty()) Q.pop();
    Q.push( Sta );
 
    P now, tem;
    while( !Q.empty() ) {
        now = Q.front();
        if(now.x == End.x && now.y == End.y) {
            return now.step;
        }
        Q.pop();
 
        for(int i=0; i<8; i++) {
            tem.x = now.x + road[i][0];
            tem.y = now.y + road[i][1];
 
            if( vis[tem.x][tem.y] || mis[tem.x][tem.y] || a[tem.x][tem.y] == '#' || !judge(tem ) )
                continue;
 
            if(0 == i || 2 == i || 5 == i || 7 == i) {
                if(!wtwalk(i, tem))
                    continue;
            }
        //   cout << tem.x+1 << " " << tem.y+1 << endl;
            force[tem.x][tem.y] = 1;
            tem.step = now.step+1;
            Q.push( tem );
        }
    }
    return -1;
}
intmain()
{
     int T;
     cin >> T;
     for(int t=1; t<=T; t++) {
          cin >> n >> m >> k;
          memset(mis, 0, sizeof(mis));
 
          for(int i=0; i<n; i++)
              cin >> a[i];
 
          P now;
          cin >> now.x >> now.y;
          now.x --, now.y --;
          now.step = 0;
 
          int years = 0;
          for(int i=1; i<=k; i++) {
             cin >> score[i].x >> score[i].y;
             score[i].x --;
             score[i].y --;
             if(score[i].x == now.x && score[i].y == now.y && i!=1){
                 years = -1;
             }
             if(i != 1) {
                mis[score[i].x][score[i].y] = 1;
             }
          }
          if(ans == -1) {
            cout << "-1" << endl;
            continue;
          }
 
          for(int i=1; i<=k; i++) {
 
             mis[score[i].x][score[i].y] = 0;
             int temp = bfs(now, score[i]);
 
             if(temp >= 0) {
                score[i].step = 0;
                now = score[i];
                years += temp;
             }
             else {
                years = -1;
                break;
             }
          }
          cout << ans << endl;
     }
     return 0;
}