Codeforces gym 101102 D 单调栈

Rectangles

time limit per test

5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Given an R×C grid with each cell containing an integer, find the number of subrectangles in this grid that contain only one distinct integer; this means every cell in a subrectangle contains the same integer.

A subrectangle is defined by two cells: the top left cell (r1, c1), and the bottom-right cell (r2, c2) (1 ≤ r1 ≤ r2 ≤ R) (1 ≤ c1 ≤ c2 ≤ C), assuming that rows are numbered from top to bottom and columns are numbered from left to right.

Input

The first line of input contains a single integer T, the number of test cases.

The first line of each test case contains two integers R and C (1 ≤ R, C ≤ 1000), the number of rows and the number of columns of the grid, respectively.

Each of the next R lines contains C integers between 1 and 109, representing the values in the row.

Output

For each test case, print the answer on a single line.

Example

input

13 33 3 13 3 12 2 5

output

16

题意：求有多少子矩阵  元素是唯一的

题解：枚举矩阵的右下角  用单调栈维护左上角可能的点

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<deque>using namespace std;typedef long long ll;int a[1005][1005],up[1005][1005];deque<int>sp;int main(){int t;scanf("%d",&t);while(t--){int n,m,i,j,k;scanf("%d%d",&n,&m);for(i=1;i<=n;i++){for(j=1;j<=m;j++){scanf("%d",&a[i][j]);up[i][j]=(a[i][j]==a[i-1][j])?up[i-1][j]:i;}}ll ans=0,sum;for(i=1;i<=n;i++){for(j=1;j<=m;j++){int p=i-up[i][j]+1;if(a[i][j]!=a[i][j-1]){while(!sp.empty())sp.pop_front();sum=0;sp.push_back(p);sum+=p;ans+=sum;}else{int now=0;while(!sp.empty()&&sp.back()>p){now++;sum-=sp.back();sp.pop_back();}for(k=1;k<=now+1;k++){sum+=p;sp.push_back(p);}ans+=sum;}}}printf("%lld\n",ans);}return 0;}