[Codeforces 1586] Omkar and Determination | 思维前缀和

Posted 2021-11-07 PushyTao

链接

description

The problem statement looms below, filling you with determination.

Consider a grid in which some cells are empty and some cells are filled. Call a cell in this grid exitable if, starting at that cell, you can exit the grid by moving up and left through only empty cells. This includes the cell itself, so all filled in cells are not exitable. Note that you can exit the grid from any leftmost empty cell (cell in the first column) by going left, and from any topmost empty cell (cell in the first row) by going up.

Let’s call a grid determinable if, given only which cells are exitable, we can exactly determine which cells are filled in and which aren’t.

You are given a grid a of dimensions n×m , i. e. a grid with n rows and m columns. You need to answer q queries $(1\le q\le 2\cdot 10^5)$. Each query gives two integers $x1,x2(1\le x1\le x2\le m)$ and asks whether the subgrid of a consisting of the columns $x1,x1+1,\dots ,x2-1,x2$ is determinable.

Input
The first line contains two integers $n,m(1\le n,m\le 10^6,n\ast m\le 10^6)$ — the dimensions of the grid a.

n lines follow. The y-th line contains m characters, the x-th of which is ‘X’ if the cell on the intersection of the the y-th row and x-th column is filled and “.” if it is empty.

The next line contains a single integer q $(1\le q\le 2\cdot 10^5)$ — the number of queries.

q lines follow. Each line contains two integers x1 and x2 $(1\le x1\le x2\le m)$, representing a query asking whether the subgrid of a containing the columns $x1,x1+1,\dots ,x2-1,x2$ is determinable.

Output
For each query, output one line containing “YES” if the subgrid specified by the query is determinable and “NO” otherwise. The output is case insensitive (so “yEs” and “No” will also be accepted).

Example
inputCopy

4 5
..XXX
...X.
...X.
...X.
5
1 3
3 3
4 5
5 5
1 5

outputCopy

YES
YES
NO
YES
NO

Note
For each query of the example, the corresponding subgrid is displayed twice below: first in its input format, then with each cell marked as “E” if it is exitable and “N” otherwise.

For the first query:

…X EEN
… EEE
… EEE
… EEE

For the second query:

X N
. E
. E
. E
Note that you can exit the grid by going left from any leftmost cell (or up from any topmost cell); you do not need to reach the top left corner cell to exit the grid.

For the third query:

XX NN
X. NN
X. NN
X. NN
This subgrid cannot be determined only from whether each cell is exitable, because the below grid produces the above “exitability grid” as well:

XX
XX
XX
XX

For the fourth query:

X N
. E
. E
. E

For the fifth query:

…XXX EENNN
…X. EEENN
…X. EEENN
…X. EEENN
This query is simply the entire grid. It cannot be determined only from whether each cell is exitable because the below grid produces the above “exitability grid” as well:

…XXX
…XX
…XX
…XX

题意

给定一个$n\ast m$的方格，在这个方格中有一些点被标记为${}^{\prime }{.}^{\prime }$，说明这个点是没有障碍的，而${}^{\prime }{X}^{\prime }$代表这个点是有障碍的，不能通过这个点，对于每个点，只能向上或者是向左走。如所说有的${}^{\prime }{.}^{\prime }$点不能走出去，那么这样的${}^{\prime }{.}^{\prime }$点就不是$exitable$
问题是：给定一个矩阵里面所有的$exitable$点，如果给出的矩阵能够唯一确定，就是$YES$，否则输出$NO$

所以问题就变成了给定的矩阵范围中有没有是${}^{\prime }{.}^{\prime }$但是不能够$exitable$的点，如果有就无法唯一确定(YES)，反之就可以唯一确定(NO)
数据范围太大，可以开$vector$来模拟二维数组，也可以使用一维数组转化

vector<int> vet[maxn];
vector<int> vis[maxn];
int sum[maxn];
int n,m;
int main() {
	cin >> n >> m;
	getchar();
	for(int i=1;i<=n;i++) vis[i].clear(),vet[i].clear(),vet[i].push_back(0);
	for(int i=0;i<=m;i++) vet[0].push_back(0);
	for(int i=1;i<=n;i++) {
		for(int j=1;j<=m;j++) {
			char c = getchar();
			int x = 0;
			if(c == 'X') x = 1;
			vet[i].push_back(x);
		}
		getchar();
		// getchar();
	}
	for(int i=1;i<=n;i++) {
		vis[i].push_back(0);
		for(int j=1;j<=m;j++) {
			if(vet[i-1][j] && vet[i][j-1]) vis[i].push_back(1);/// ???
			else vis[i].push_back(0);
		}
	}
	for(int j=1;j<=m;j++) {
		for(int i=1;i<=n;i++) {
			sum[j] += vis[i][j];
		}
		sum[j] += sum[j-1];
		// cout << sum[j] << ' ';
	}
	// puts("\\n input");
	int t;
	cin >> t;
	while(t --) {
		int l = read,r = read;
		if(sum[r] - sum[l]) puts("NO");
		else puts("YES");
	}
	return 0;
}
/**


**/