标签：des style blog color os io strong for ar
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:
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 Tth minute. Its status is considered during the change at the beginning of the Tth 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 :)
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 Ti, Xi, Yi, They mean the minion who is located in (Xi, Yi) will leave the game at the end of the Tith minutes. 1<=Ti<= F, 1<=Xi<=N, 1<=Yi<=M.
For each case, output N lines as a matrix which shows the status of each minion after F minutes.
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
010 1X0 010 0000X 11000 00X00 X0000 00000
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 */
ZOJ Problem Set - 3804 YY's Minions
标签：des style blog color os io strong for ar
原文地址：http://www.cnblogs.com/cancangood/p/3935670.html
