CF 548B Mike and Fun

B. Mike and Fun

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Mike and some bears are playing a game just for fun. Mike is the judge. All bears except Mike are standing in ann × m grid, there's exactly one bear in each cell. We denote the bear standing in column number j of row number iby (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.

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 and1 ≤ 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.

Sample test(s)

input

5 4 5
0 1 1 0
1 0 0 1
0 1 1 0
1 0 0 1
0 0 0 0
1 1
1 4
1 1
4 2
4 3

output

3
4
3

#include <algorithm>
#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
#include <map>

using namespace std;
const int N=5E2+5;
int a[N][N];
int fans;
int n,m,q,x,y,cur,ans[N];
int main()
{
    cin>>n>>m>>q;
    for (int i = 1 ; i <= n;i++ )
    {
        for (int j = 1; j <= m ; j++ )
                scanf("%d",&a[i][j]);
        cur = 0;
        for (int j = 1; j <=m ;j++ )
            if (a[i][j]==1)
            {
                cur++;
                ans[i]=max(cur,ans[i]);
            }
            else
            {
               cur = 0;
            }
    }
    for ( int i = 1 ; i <= q; i++ )
    {
        scanf("%d %d",&x,&y);
        a[x][y]=a[x][y]^1;
       // if (i==3) cout<<a[x][y]<<"sadsadasd"<<endl;
        cur = 0;
        ans[x]=0;
        for (int j = 1; j <=m ;j++ )
            if (a[x][j]==1)
            {
                cur++;
                ans[x]=max(cur,ans[x]);
            }
            else
            {
               cur = 0;
            }
        fans=-1;
        for (int j = 1;j <= n ; j++ )
            if (ans[j]>fans)
            {
                fans=ans[j];
            }
        cout<<fans<<endl;
    }


    return 0;
}

3
4

比赛的时候不懂为什么就没做出来....
其实很容易想到一个o(q*(n+m))的做法...
就是每次更新，要同时更新当前更新行的最大连续和....O（m）可以完成...然后在O（n）扫一遍，找到所有行中的最大值。
然后需要注意的是，在第一次更改之前就要把每个行的最大值处理出来l..
然后cf机器真是够快，O(n*m*q)的1.2S过。。。。