HDU 2612 find a way

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 2612

Q: It takes the shortest time for Y and m to go to the same KFC (there may be multiple KFC instances)

My method is to search for the sum of the shortest time of two people from KFC. The only thing that hurts is the first few of TLE's. Yes.

/**     Danceonly*/#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>using namespace std;typedef long long LL;#define MAX(a,b) (a > b ? a : b)#define MIN(a,b) (a < b ? a : b)const int maxn = 10000007;struct node{      int x;      int y;}kfc[100005];struct abc{      int x,y;      int step;}q[100005],e;int maze[205][205],vis[205][205];int dir[4][2] = {1,0,0,1,-1,0,0,-1};int N,M;int BFS(int sx,int sy,int A){      int tail,head,step,x,y;      head = tail = 0;      q[head].x = sx;      q[head].y = sy;      q[head].step = 0;      vis[sx][sy] = 1;      while (head <= tail)      {            x = q[head].x;            y = q[head].y;            step = q[head].step;            for (int i=0;i<4;i++)            {                  e.x = x + dir[i][0];                  e.y = y + dir[i][1];                  if (maze[e.x][e.y] ==  0 || vis[e.x][e.y])continue ;                  e.step = step + 1;                  vis[e.x][e.y] = 1;                  if (maze[e.x][e.y] ==  A)                  {                        return e.step;                  }                  q[++tail] = e;            }            head++;      }      return maxn;}void show(){      for (int i=1;i<=N;i++)      {            for (int j=1;j<=M;j++)                  printf("%d ",maze[i][j]);            printf("\n");      }}void init(){      memset(vis,0,sizeof(vis));      memset(maze,0,sizeof(maze));      return ;}int main(){      while (scanf("%d%d",&N,&M) == 2)      {            int Yx,Yy,Mx,My;            memset(maze,0,sizeof(maze));            int c = 0;            for (int i=1;i<=N;i++)            {                  char ss[1005];                  scanf("%s",ss);                  for (int j=0;j<M;j++)                        if (ss[j] == '.')maze[i][j+1] = 1;                        else if (ss[j] == '@'){kfc[c].x = i;kfc[c++].y = j + 1;maze[i][j+1] = 1;}                        else if (ss[j] == 'Y'){maze[i][j+1] = 2;Yx = i;Yy = j + 1;}                        else if (ss[j] == 'M'){maze[i][j+1] = 3;Mx = i;My = j + 1;}            }            //show();            int ans = maxn;            for (int i=0;i<c;i++)            {                  int sum = 0;                  memset(vis,0,sizeof(vis));                  sum += BFS(kfc[i].x,kfc[i].y,2);                  if (sum + abs(kfc[i].x - Mx) + abs(kfc[i].y-My) >= ans)                        continue ;                  memset(vis,0,sizeof(vis));                  sum += BFS(kfc[i].x,kfc[i].y,3);                  if (ans > sum)ans = sum;            }            printf("%d\n",ans * 11);      }      return 0;}/*4 4Y.#@.....#..@..M4 4Y.#@.....#..@#.M5 5Y..@..#....#...@..M.#...#*/