ZOJ Problem Set - 3804 YY's Minions

学习：换一个角度考虑问题。YY's Minions

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Despite YY's so much homework, she would like to take some time to play with her minions first.

YY lines her minions up to an N*M matrix. Every minion has two statuses: awake or asleep. We use 0(the digit) to represent that it is asleep, and 1 for awake. Also, we define the minions who are around a minion closest in one of the eight directions its neighbors. And every minute every minion will change its status by the following specific rules:

• If this minion is awake, and the number of its neighbors who are awake is less than 2, this minion will feel lonely and turn to asleep.
• If this minion is awake, and the number of its neighbors who are awake is more than 3, this minion will turn to asleep for it will feel too crowded.
• If this minion is awake, and the number of its neighbors who are awake is exactly 2 or 3, this minion will keep being awake and feel very happy.
• If this minion is asleep, and the number of its neighbors who are awake is exactly 3, this minion will wake up because of the noise. Note that all changes take place at the same time at the beginning of a specific minute.

   

  Also, some minions will get bored and leave this silly game. We use 'X's to describe them. We suppose that a minion would leave after T minutes. It will leave at the end of the T^th minute. Its status is considered during the change at the beginning of the T^th minute, and should be ignored after that. Of course, one minion will not leave twice!

  YY is a girl full of curiosity and wants to know every minion's status after F minutes. But you know she is weak and lazy! Please help this cute girl to solve this problem :)

  Input

  There are multiple test cases.

  The first line contains the number of test cases Q. 1<=Q<=100.
  For each case, there are several lines:
  The first line contains four integers N, M, F, K. K means the number of leaving messages. 1<=N, M<=50, 1<=F<=1000, 1<=K<=N*M.
  Next N lines are the matrix which shows the initial status of each minion. Each line contains M chars. We guarantee that 'X' wouldn't appear in initial status matrix.
  And next K lines are the leaving messages. Each line contains three integers T_i, X_i, Y_i, They mean the minion who is located in (X_i, Y_i) will leave the game at the end of the T_i^th minutes. 1<=T_i<= F, 1<=X_i<=N, 1<=Y_i<=M.

  Output

  For each case, output N lines as a matrix which shows the status of each minion after F minutes.

  Sample Input

  2
  3 3 2 1
  101
  110
  001
  1 2 2
  5 5 6 3
  10111
  01000
  00000
  01100
  10000
  2 3 3
  2 4 1
  5 1 5
  

  Sample Output

  010
  1X0
  010
  0000X
  11000
  00X00
  X0000
  00000
  

  Hint

  For case 1:
  
  T=0, the game starts
  101
  110
  001
  ---------------
  at the beginning of T=1, a change took place
  100
  101
  010
  ---------------
  at the end of T=1 (the minion in (2,2) left)
  100
  1X1
  010
  ---------------
  at the beginning of T=2, a change took place
  010
  1X0
  010
  ---------------
  at the end of T=2 (nothing changed for no minion left at T=2)
  010
  1X0
  010
  

  #include<iostream>
  #include<cstring>
  #include<cstdio>
  #include<algorithm>
  using namespace std;
  char str[55][55];
  int a[100][100],aa[100][100];
  int t[2550],x[2550],y[2550];
  int d[8][2]= {{1,0},{1,1},{1,-1},{0,1},{0,-1},{-1,0},{-1,1},{-1,-1}};
  int main()
  {
      int q,n,m,f,k,r,e;
      int i,j,cnt,c;
      scanf("%d",&q);
      while(q--)
      {
          memset(a,0,sizeof(a));
          memset(t,0,sizeof(t));
          memset(x,0,sizeof(x));
          memset(y,0,sizeof(y));
          memset(str,0,sizeof(str));
          scanf("%d%d%d%d",&n,&m,&f,&k);
          getchar();
          for(i=0; i<n; i++)
          {
              for(j=0; j<m; j++)
                  scanf("%c",&str[i][j]);
              getchar();
          }
  
          for(i=0; i<n; i++)
          {
              for(j=0; j<m; j++)
              {
                  if(str[i][j]=='1')
                      a[i][j]=1;
                  if(str[i][j]=='0')
                      a[i][j]=0;
              }
          }
           
          for(i=0; i<k; i++)
              scanf("%d%d%d",&t[i],&x[i],&y[i]);
          for(r=1; r<=f; r++)
          {
              memset(aa,0,sizeof(aa));
              for(i=0; i<n; i++)
              {
                  for(j=0; j<m; j++)
                  {
                      if(a[i][j]==1)
                      {
                          for(e=0;e<8;e++)
                          {
                              int xx=i+d[e][0];
                              int  yy=j+d[e][1];
                              if(xx>=0&&yy>=0&&xx<n&&yy<m)
                                  aa[xx][yy]++;
                          }
                      }
                  }
              }
              for(i=0; i<n; i++)
              {
                  for(j=0; j<m; j++)
                  {
                      if(a[i][j]==2)
                        continue;
                      if(a[i][j]==1)
                      {
                          if(aa[i][j]<2||aa[i][j]>3)
                              a[i][j]=0;
                      }
                      else if(a[i][j]==0)
                      {
                          if(aa[i][j]==3)
                              a[i][j]=1;
                      }
                  }
              }
              for(i=0;i<k;i++)
              {
                  if(r==t[i])
                      a[x[i]-1][y[i]-1]=2;
              }
          }
          for(i=0; i<n; i++)
          {
              for(j=0; j<m; j++)
              {
                  if(a[i][j]==2)
                      printf("X");
                  else
                      printf("%d",a[i][j]);
              }
              printf("\n");
          }
      }
      return 0;
  }
  /*
  3
  5 5 6 5
  10111
  01000
  00000
  01100
  10000
  2 3 3
  2 4 1
  5 1 5
  4 3 3
  1 5 5
  
  */