玲珑比赛 A -- chess play【记忆模拟】_a

本文链接：https://blog.youkuaiyun.com/DTL66/article/details/53056835

本文介绍了一道玲珑杯ACM比赛中关于棋盘操作的问题，包括添加黑白棋子及交换行的操作，并提供了完整的AC代码实现。

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玲珑杯”ACM比赛 Round #4

Start Time：2016-11-05 12:00:00 End Time：2016-11-05 17:00:00 Refresh Time：2016-11-06 21:38:47 Private

A -- chess play

Time Limit：10s Memory Limit：64MByte

Submissions：777Solved：266

DESCRIPTION

 Bob has a n×mn×m chessboard, which has two kinds of chess pieces.one is black, and the other is white.
 As for chessboard, you can do three operations：
1 1 x y1 1 x y (means add a white chess piece to position (x, y))
1 2 x y1 2 x y(means add a black chess piece to position (x, y))
2 x1 x22 x1 x2(means swap the x1x1 row and x2x2 row)
 (note: if you add one new chess to a position where there already is a chess, you need throw the old chess away firstly.)
 In the initial time the chessboard is empty, and your mission is to show the final chessboard.
 Please look at the example for more details.

INPUT

   There are multiple test cases.The first line is a number T (
   
   T ≤10T ≤10), which means the number of cases.For each case, the first line contains three integers n, m and q 
   
   (1≤n,m≤103,1≤q≤106)(1≤n,m≤103,1≤q≤106). Then follow 
   
   qq lines are the operations.As for each operation, 
    
   1≤xi,yi≤n1≤xi,yi≤n

OUTPUT

   Print n lines which containing m characters - the final chessboard.
  

SAMPLE INPUT

    22 2 21 1 1 12 1 24 3 41 1 1 12 1 21 2 1 12 2 3
   

SAMPLE OUTPUT

    ..w.b.....w.....
   

AC代码：

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;
 
const int MAXN=1e3+11;
char G[MAXN][MAXN]; 
int a[MAXN];

int main()
{
	int T,N,M,Q; scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d%d",&N,&M,&Q);
		for(int i=0;i<N;++i) {
			for(int j=0;j<M;++j) G[i][j]='.';
			G[i][M]='\0'; a[i]=i;
		}
		while(Q--) {
		     int x,y,o1,o2;
			 scanf("%d",&o1);
			 if(o1==1) {
                   scanf("%d%d%d",&o2,&x,&y);
				   G[a[x-1]][y-1]=o2==1?'w':'b';
             }
			 else {
			 	scanf("%d%d",&x,&y);
			 	swap(a[x-1],a[y-1]);
			 }
		}
		for(int i=0;i<N;++i) printf("%s\n",G[a[i]]);
    } 
	return 0;
 }