Codeforces Round #305 (Div. 2) B.Mike and Fun

最新推荐文章于 2024-08-28 22:49:31 发布

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最新推荐文章于 2024-08-28 22:49:31 发布

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本文介绍了一个关于游戏状态更新及得分计算的问题。玩家需要在给定的n*m矩阵上改变特定位置的状态，并计算每行中连续相同状态的最大数量。通过输入输出示例展示了具体的实现过程。

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Mike and some bears are playing a game just for fun. Mike is the judge. All bears except Mike are standing in an n × m grid, there's exactly one bear in each cell. We denote the bear standing in column number j of row number i by (i, j). Mike's hands are on his ears (since he's the judge) and each bear standing in the grid has hands either on his mouth or his eyes.

$\text{[math]}$

They play for q rounds. In each round, Mike chooses a bear (i, j) and tells him to change his state i. e. if his hands are on his mouth, then he'll put his hands on his eyes or he'll put his hands on his mouth otherwise. After that, Mike wants to know the score of the bears.

Score of the bears is the maximum over all rows of number of consecutive bears with hands on their eyes in that row.

Since bears are lazy, Mike asked you for help. For each round, tell him the score of these bears after changing the state of a bear selected in that round.

Input

The first line of input contains three integers n, m and q (1 ≤ n, m ≤ 500 and 1 ≤ q ≤ 5000).

The next n lines contain the grid description. There are m integers separated by spaces in each line. Each of these numbers is either 0(for mouth) or 1 (for eyes).

The next q lines contain the information about the rounds. Each of them contains two integers i and j (1 ≤ i ≤ n and 1 ≤ j ≤ m), the row number and the column number of the bear changing his state.

Output

After each round, print the current score of the bears.

题意：给出一个 n * m 的矩阵，q 表示为游戏的次数。

输入矩阵后，每给出两个数代表在矩阵所在的点，如果是 1 的话变为 0 ，如果是 0 的话变为 1 。

然后判断矩阵中每行最多出现连续 1 的数量情况。（要用scanf , printf 进行输入输出会比用cin , cout 输入输出更省时，不会TL）

#include<cstdio>
#include<cstring>
#include<iostream>
#include<utility>
#include<string>
#include<vector>
#include<algorithm>
#include<queue>
#include<cstdlib>
#include<cmath>
#include<stack>

using namespace std;

const int inf = 0x3f3f3f3f ;
int a[505][505] ;

int main()
{
	int n , m , q ;
	scanf("%d%d%d",&n,&m,&q);
	int i , j ;
	for(i = 1 ; i <= n ; ++i)
	{
		for(j = 1 ; j <= m ; ++j)
			scanf("%d",&a[i][j]);
	}
	while(q--)
	{
		int b , c ,ans = 0 , maxx=-inf ;
		scanf("%d%d",&b , &c);
		if(!a[b][c])
			a[b][c] = 1 ;
		else
			a[b][c] = 0 ;
		for(i = 1 ; i <= n ; ++i)
		{
			ans = 0 ;
			for(j = 1 ; j <= m ; ++j)
			{
				if(a[i][j])
					ans++ ;
				else
					ans = 0 ;
				maxx = max(maxx , ans);
			}
		}
		printf("%d\n",maxx);
	}

	return 0;
}