ZOJ Problem Set - 3804 YY's Minions

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cmath>
  4 #include <vector>
  5 #include <cstring>
  6 #include <string>
  7 #include <algorithm>
  8 #include <string>
  9 #include <set>
 10 #include <functional>
 11 #include <numeric>
 12 #include <sstream>
 13 #include <stack>
 14 #include <map>
 15 #include <queue>
 16 
 17 #define CL(arr, val)    memset(arr, val, sizeof(arr))
 18 
 19 #define ll long long
 20 #define inf 0x7f7f7f7f
 21 #define lc l,m,rt<<1
 22 #define rc m + 1,r,rt<<1|1
 23 #define pi acos(-1.0)
 24 
 25 #define L(x)    (x) << 1
 26 #define R(x)    (x) << 1 | 1
 27 #define MID(l, r)   (l + r) >> 1
 28 #define Min(x, y)   (x) < (y) ? (x) : (y)
 29 #define Max(x, y)   (x) < (y) ? (y) : (x)
 30 #define E(x)        (1 << (x))
 31 #define iabs(x)     (x) < 0 ? -(x) : (x)
 32 #define OUT(x)  printf("%I64d\n", x)
 33 #define lowbit(x)   (x)&(-x)
 34 #define Read()  freopen("a.txt", "r", stdin)
 35 #define Write() freopen("b.txt", "w", stdout);
 36 #define maxn 1000000000
 37 #define N 55
 38 using namespace std;
 39 
 40 int n,m,f,k;
 41 char A[N][N],B[N][N];
 42 int dir[8][2]={{-1,0},{-1,1},{-1,-1},{0,1},{0,-1},{1,0},{1,1},{1,-1}};
 43 struct point
 44 {
 45     int x,y,t;
 46     bool operator < (const point &a) const
 47     {
 48         return t<a.t;
 49     }
 50 }p[N*N];
 51 
 52 void solve()
 53 {
 54     for(int i=0;i<n;i++)
 55     {
 56         for(int j=0;j<m;j++)
 57         {
 58             B[i][j]=A[i][j];
 59             int ans=0;
 60             for(int k=0;k<8;k++)
 61             {
 62                 int x=i+dir[k][0];
 63                 int y=j+dir[k][1];
 64                 if(x<0||x>=n||y<0||y>=m) continue;
 65                 if(A[x][y]==‘1‘) ans++;
 66             }
 67             if(A[i][j]==‘1‘)
 68             {
 69                 if(ans<2) B[i][j]=‘0‘;
 70                 else if(ans>3) B[i][j]=‘0‘;
 71                 else B[i][j]=‘1‘;
 72             }
 73             else if(A[i][j]==‘0‘)
 74             {
 75                 if(ans==3) B[i][j]=‘1‘;
 76             }
 77         }
 78     }
 79     for(int i=0;i<n;i++)
 80         for(int j=0;j<m;j++)
 81         A[i][j]=B[i][j];
 82 }
 83 int main()
 84 {
 85    //Read();
 86    //Write()
 87     int q;
 88     scanf("%d",&q);
 89     while(q--)
 90     {
 91         scanf("%d%d%d%d",&n,&m,&f,&k);
 92         getchar();
 93         for(int i=0;i<n;i++) scanf("%s",A[i]);
 94         for(int i=0;i<k;i++)
 95         {
 96             scanf("%d%d%d",&p[i].t,&p[i].x,&p[i].y);
 97             p[i].x--;
 98             p[i].y--;
 99         }
100         sort(p,p+k);
101         for(int i=1,j=0;i<=f;i++)
102         {
103             solve();
104             while(p[j].t==i)
105             {
106                 A[p[j].x][p[j].y]=‘X‘;
107                 j++;
108             }
109         }
110         for(int i=0;i<n;i++)
111             printf("%s\n",A[i]);
112     }
113    return 0;
114 }