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.
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.
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 3 4
题意：
给出一个n*m的0，1矩阵
然后有k次操作，每次操作给出一个坐标，代表对该坐标的数字取反
每次操作之后，输出矩阵里每一行连续1的个数最大的那个值
思路：
由于数据和查询都不是很大，可以直接暴力解决
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;
#define LS 2*i
#define RS 2*i+1
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 500005
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define lowbit(x) (x&-x)
struct node
{
int num[505],sum,maxn;
} a[505];
int main()
{
int n,m,i,j,k;
scanf("%d%d%d",&n,&m,&k);
memset(a,0,sizeof(a));
for(i = 1; i<=n; i++)
{
a[i].maxn=0;
a[i].sum=0;
for(j = 1; j<=m; j++)
{
scanf("%d",&a[i].num[j]);
if(a[i].num[j]==1)
a[i].sum+=a[i].num[j];
else
{
a[i].maxn = max(a[i].maxn,a[i].sum);
a[i].sum = 0;
}
}
a[i].maxn = max(a[i].maxn,a[i].sum);
}
while(k--)
{
int x,y;
scanf("%d%d",&x,&y);
if(a[x].num[y]==1)
{
a[x].num[y] = 0;
}
else
{
a[x].num[y] = 1;
}
a[x].maxn=0;
a[x].sum=0;
for(j = 1; j<=m; j++)
{
if(a[x].num[j]==1)
a[x].sum+=a[x].num[j];
else
{
a[x].maxn = max(a[x].maxn,a[x].sum);
a[x].sum = 0;
}
}
a[x].maxn = max(a[x].maxn,a[x].sum);
int maxn = -1;
for(i = 1; i<=n; i++)
maxn=max(maxn,a[i].maxn);
printf("%d\n",maxn);
}
return 0;
}