YY’s MinionsTime Limit: 2 Seconds Memory Limit: 65536 KB
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 minionclosest 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 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 afterT 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 🙂
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 containsM chars. We guarantee that ‘X’ wouldn’t appear in initial status matrix.And next K lines are the leaving messages. Each line contains three integersTi, 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.
Output
For each case, output N lines as a matrix which shows the status of each minion afterF minutes.
Sample Input23 3 2 11011100011 2 25 5 6 310111010000000001100100002 3 32 4 15 1 5Sample Output0101X00100000X1100000X00X000000000HintFor case 1:T=0, the game starts101110001—————at the beginning of T=1, a change took place100101010—————at the end of T=1 (the minion in (2,2) left)1001X1010—————at the beginning of T=2, a change took place0101X0010—————at the end of T=2 (nothing changed for no minion left at T=2)0101X0010
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题意：一个矩阵中有 n*m个宠物，每一个宠物都有一个状态， 1代表醒着的，0代表睡着的， ’X‘代表的是将要离开的，如果这个宠物（醒着的）的周围醒着的个数>3 || <2它就会睡着， 如果这个宠物（睡着的）的周围醒着的个数==3就会醒来， 每一分钟都会有变换一个状态！ 其中会有些宠物会在给定的时间内离开。求解给定t分钟后，所有宠物的状态【解题思路】：DFS 搜索八个方向，然后模拟一下，注意要注意预处理的dir数组，看清题目要求，动物睡眠状态跟周围的动物数目关系。做完之后感觉就是，搜索部分容易实现，但是如果有多个宠物将要离开，而且在不同的时间段，而且每个时间段，都会影响其周围的宠物的状态！这个部分就要仔细一点了，既然多个时间段，我们把所有时间段从小大大排列一遍！，依次处理，同步DFS搜索，，更新状态！，具体看代码了
代码：
#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;char mapp[1005][1005];char runq[1005][1005];int dir[8][2]= {0,-1,-1,-1,-1,0,-1,1,0,1,1,1,1,0,1,-1};//这里数组开成[8][4],WA了一次！int N,M,F,K;int t,t1,x1,y1;int i,j,tot=1;struct node{int t,b,xx;bool operator <(const node &c) const{return t<c.t;}} runway[1005];bool judge(int x,int y){if(x>0&&x<N&&y>0&&y<M) return true;else return false;}void dfs(){int awake=0,asleep;for(int i=0; i<N; i++)for(int j=0; j<M; j++){int maxx=0;for(int p=0; p<8; p++){int dx=i+dir[p][0];int dy=j+dir[p][1];if(dx<0||dx>=N||dy<0||dy>=M) continue;if(mapp[dx][dy]=='1') maxx++;//if(judge(dx,dy))//{//if(mapp[dx][dy]=='1') maxx++;//}}runq[i][j]=mapp[i][j];if(mapp[i][j]=='1'){if(maxx>3||maxx<2) runq[i][j]='0';else runq[i][j]='1';}else if(mapp[i][j]=='0'){if(maxx==3) runq[i][j]='1';}}for(int i=0; i<N; i++)for(int j=0; j<M; j++){mapp[i][j]=runq[i][j];}}int main(){cin>>t;while(t–){cin>>N>>M>>F>>K;for(i=0; i<N; i++)scanf("%s",mapp[i]);for(i=0; i<K; i++){cin>>runway[i].t>>runway[i].b>>runway[i].xx;runway[i].b–;runway[i].xx–;}sort(runway,runway+K);for(i=1,j=0; i<=F; i++) //注意动态更新！{dfs();while(j<K&&runway[j].t==i){mapp[runway[j].b][runway[j].xx]='X';j++;}}for(i=0; i<N; i++)printf("%s\n",mapp[i]);}return 0;}
积极的人在每一次忧患中都看到一个机会，
