Gym - 101102D Rectangles （单调栈）

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

1
3 3
3 3 1
3 3 1
2 2 5

Output

16

题意:问由单一数字组成的矩阵的个数。
思路：
dp[i][j]表示以位置i,j为右下角的矩阵个数。
先处理出当前位置向上延伸相同的数字最高有多高。用num[i]记录。
用单调栈处理出在此之前的，第一个小于自己的num[i].
判断中间有没有插入别的数，再进行处理。详见solve函数。

 #include<iostream>
 #include<algorithm>
 #include<vector>
 #include<stack>
 #include<queue>
 #include<map>
 #include<set>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<ctime>
 #define fuck(x) cout<<#x<<" = "<<x<<endl;
 #define debug(a,i) cout<<#a<<"["<<i<<"] = "<<a[i]<<endl;
 #define ls (t<<1)
 #define rs ((t<<1)|1)
 using namespace std;
 typedef long long ll;
 typedef unsigned long long ull;
 const int maxn = ;
 const int maxm = ;
 const int inf = 2.1e9;
 const ll Inf = ;
 const int mod = ;
 const double eps = 1e-;
 const double pi = acos(-);
 int n,m;
 int mp[maxn][maxn];
 int num[maxn];
 ll ans;
 struct  node{
     int num,pos;
 };
 stack<node>st;
 int pre[maxn];
 int pre1[maxn];
 ll dp[maxn];
 void solve(int t){
     memset(dp,,sizeof(dp));
     memset(pre,,sizeof(pre));
     while(!st.empty()){
         st.pop();
     }
     num[]=-;
     for(int i=m;i>=;i--){
         while(!st.empty()&&st.top().num>num[i]){
             pre[st.top().pos]=i;
             st.pop();
         }
         st.push(node{num[i],i});
     }
     int k=;
     for(int i=;i<=m;i++){
         if(mp[t][i]!=mp[t][i-]){
             k=i;
         }
         pre1[i]=k;
     }
     int pree;
     for(int i=;i<=m;i++){
         if(pre1[i]>pre[i]){
             dp[i]=1ll*num[i]*(i-pre1[i]+);
             ans+=dp[i];
         }
         else{
             dp[i]=1ll*num[i]*(i-pre[i])+dp[pre[i]];
             ans+=dp[i];
         }
     }
 }

 int main()
 {
     int T;
     scanf("%d",&T);
     while(T--){
         scanf("%d%d",&n,&m);
         for(int i=;i<=n;i++){
             for(int j=;j<=m;j++){
                 scanf("%d",&mp[i][j]);
             }
         }
         ans=;
         memset(num,,sizeof(num));
         for(int i=;i<=n;i++){
             for(int j=;j<=m;j++){
                 if(mp[i][j]==mp[i-][j]){
                     num[j]++;
                 }
                 else{
                     num[j]=;
                 }
             }
             solve(i);
         }
         printf("%lld\n",ans);
     }
     return ;
 }