题目连接：
Description
Nanami is an expert at playing games. This day, Nanami's good friend Hajime invited her to watch a game of baseball. Unwilling as she was, she followed him to the stadium. But Nanami had no interest in the game, so she looked around to see if there was something that might interest her. That's when she saw the digital board at one end of the stadium.
The digital board is n pixels in height and m pixels in width, every pixel is either light or dark. The pixels are described by its coordinate. The j-th pixel of the i-th line is pixel (i, j). The board displays messages by switching a combination of pixels to light, and the rest to dark. Nanami notices that the state of the pixels on the board changes from time to time. At certain times, certain pixels on the board may switch from light to dark, or from dark to light.
Nanami wonders, what is the area of the biggest light block such that a specific pixel is on its side. A light block is a sub-rectangle of the board, in which all pixels are light. Pixel (i, j) belongs to a side of sub-rectangle with (x1, y1) and (x2, y2) as its upper-left and lower-right vertex if and only if it satisfies the logical condition:
((i = x1 or i = x2) and (y1 ≤ j ≤ y2)) or ((j = y1 or j = y2) and (x1 ≤ i ≤ x2)).
Nanami has all the history of changing pixels, also she has some questions of the described type, can you answer them?Input
The first line contains three space-separated integers n, m and q (1 ≤ n, m, q ≤ 1000) — the height and width of the digital board, and the number of operations.
Then follow n lines, each line containing m space-separated integers. The j-th integer of the i-th line is ai, j — the initial state of pixel (i, j).
If ai, j = 0, pixel (i, j) is initially dark.
If ai, j = 1, pixel (i, j) is initially light. Then follow q lines, each line containing three space-separated integers op, x, and y (1 ≤ op ≤ 2; 1 ≤ x ≤ n; 1 ≤ y ≤ m), describing an operation.If op = 1, the pixel at (x, y) changes its state (from light to dark or from dark to light).
If op = 2, Nanami queries the biggest light block with pixel (x, y) on its side.Output
For each query, print a single line containing one integer — the answer to Nanami's query.
Sample Input
3 4 5
0 1 1 0 1 0 0 1 0 1 1 0 2 2 2 2 1 2 1 2 2 1 2 3 2 2 2Sample Output
0
2 6Hint
题意
给你一个01矩阵，然后有两个操作，1是将x y取反，2是问以x,y为边界的最大全1矩形的面积是多少
题解：
其实就是瞎暴力……
一看数据范围，只要能做到修改操作是o(n)和查询操作是o(n)的就好了
这个直接用单调栈那种思想，直接暴力去莽一波就好了
对于每一个格子维护四个值，l[x][y],r[x][y],u[x][y],d[x][y]表示这个格子最多往四个方向延展多少
代码
#includeusing namespace std;const int maxn = 1005;int a[maxn][maxn],n,m,q,up[maxn][maxn],down[maxn][maxn],l[maxn][maxn],r[maxn][maxn];int x,y,op;void update(){ a[x][y]=1-a[x][y]; for(int j=1;j<=m;j++) if(a[x][j]) l[x][j]=l[x][j-1]+1; else l[x][j]=0; for(int j=m;j>=1;j--) if(a[x][j]) r[x][j]=r[x][j+1]+1; else r[x][j]=0; for(int i=1;i<=n;i++) if(a[i][y]) up[i][y]=up[i-1][y]+1; else up[i][y]=0; for(int i=n;i>=1;i--) if(a[i][y]) down[i][y]=down[i+1][y]+1; else down[i][y]=0;}void query(){ if(a[x][y]==0) { printf("0\n"); return; } int ans = 0; int U=1e9,D=1e9,L=1e9,R=1e9; for(int i=y;i>=1;i--) { U=min(U,up[x][i]); D=min(D,down[x][i]); ans=max(ans,(U+D-1)*(y-i+1)); } U=1e9,D=1e9,L=1e9,R=1e9; for(int i=y;i<=m;i++) { U=min(U,up[x][i]); D=min(D,down[x][i]); ans=max(ans,(U+D-1)*(i-y+1)); } U=1e9,D=1e9,L=1e9,R=1e9; for(int i=x;i>=1;i--) { L=min(L,l[i][y]); R=min(R,r[i][y]); ans=max(ans,(L+R-1)*(x-i+1)); } U=1e9,D=1e9,L=1e9,R=1e9; for(int i=x;i<=n;i++) { L=min(L,l[i][y]); R=min(R,r[i][y]); ans=max(ans,(L+R-1)*(i-x+1)); } printf("%d\n",ans);}int main(){ scanf("%d%d%d",&n,&m,&q); for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) scanf("%d",&a[i][j]); for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) if(a[i][j]) l[i][j]=l[i][j-1]+1; for(int j=m;j>=1;j--) if(a[i][j]) r[i][j]=r[i][j+1]+1; } for(int j=1;j<=m;j++) { for(int i=1;i<=n;i++) if(a[i][j]) up[i][j]=up[i-1][j]+1; for(int i=n;i>=1;i--) if(a[i][j]) down[i][j]=down[i+1][j]+1; } for(int i=1;i<=q;i++) { scanf("%d%d%d",&op,&x,&y); if(op==1)update(); else query(); }}