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.
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.
For each test case, print the answer on a single line.
1
3 3
3 3 1
3 3 1
2 2 5
16
分析：按行处理，对每个点找到向上和向左的的最大矩形；
然后对于左边小矩形，直接加上之前答案即可；
向上向左预处理后RMQ+二分找最大矩形即可，复杂度O(n2logn)；
代码：
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <climits> #include <cstring> #include <string> #include <set> #include <map> #include <queue> #include <stack> #include <vector> #include <list> #define rep(i,m,n) for(i=m;i<=n;i++) #define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++) #define mod 1000000007 #define inf 0x3f3f3f3f #define vi vector<int> #define pb push_back #define mp make_pair #define fi first #define se second #define ll long long #define pi acos(-1.0) #define pii pair<int,int> #define Lson L, mid, ls[rt] #define Rson mid+1, R, rs[rt] #define sys system("pause") #define intxt freopen("in.txt","r",stdin) const int maxn=1e3+10; using namespace std; ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);} ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;} inline ll read() { ll x=0;int f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int n,m,k,t,a[maxn][maxn],le[maxn][maxn],up[maxn][maxn],ans[maxn],mi[10][maxn],p[maxn]; void init(int now) { for(int i=1;i<=m;i++)mi[0][i]=up[now][i]; for(int i=1;i<=9;i++) for(int j=1;j+(1<<i)-1<=m;j++) mi[i][j]=min(mi[i-1][j],mi[i-1][j+(1<<(i-1))]); } int query(int l,int r) { int x=p[r-l+1]; return min(mi[x][l],mi[x][r-(1<<x)+1]); } int main() { int i,j; rep(i,2,1000)p[i]=1+p[i>>1]; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); rep(i,1,n)rep(j,1,m)a[i][j]=read(); rep(i,1,n)rep(j,1,m) { up[i][j]=le[i][j]=1; if(a[i][j]==a[i-1][j])up[i][j]=up[i-1][j]+1; if(a[i][j]==a[i][j-1])le[i][j]=le[i][j-1]+1; } ll ret=0; rep(i,1,n) { init(i); memset(ans,0,sizeof(ans)); rep(j,1,m) { int l=j-le[i][j]+1,r=j,now_ans; while(l<=r) { int mid=l+r>>1; if(query(mid,j)>=up[i][j])now_ans=mid,r=mid-1; else l=mid+1; } ans[j]+=up[i][j]*(j-now_ans+1); if(now_ans>j-le[i][j]+1)ans[j]+=ans[now_ans-1]; } rep(j,1,m)ret+=ans[j]; } printf("%lld ",ret); } //system("Pause"); return 0; }