5 seconds
256 megabytes
standard input
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.
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+;
using namespace std;
ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}
inline ll read()
{
ll x=;int f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
int n,m,k,t,a[maxn][maxn],le[maxn][maxn],up[maxn][maxn],ans[maxn],mi[][maxn],p[maxn];
void init(int now)
{
for(int i=;i<=m;i++)mi[][i]=up[now][i];
for(int i=;i<=;i++)
for(int j=;j+(<<i)-<=m;j++)
mi[i][j]=min(mi[i-][j],mi[i-][j+(<<(i-))]);
}
int query(int l,int r)
{
int x=p[r-l+];
return min(mi[x][l],mi[x][r-(<<x)+]);
}
int main()
{
int i,j;
rep(i,,)p[i]=+p[i>>];
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
rep(i,,n)rep(j,,m)a[i][j]=read();
rep(i,,n)rep(j,,m)
{
up[i][j]=le[i][j]=;
if(a[i][j]==a[i-][j])up[i][j]=up[i-][j]+;
if(a[i][j]==a[i][j-])le[i][j]=le[i][j-]+;
}
ll ret=;
rep(i,,n)
{
init(i);
memset(ans,,sizeof(ans));
rep(j,,m)
{
int l=j-le[i][j]+,r=j,now_ans;
while(l<=r)
{
int mid=l+r>>;
if(query(mid,j)>=up[i][j])now_ans=mid,r=mid-;
else l=mid+;
}
ans[j]+=up[i][j]*(j-now_ans+);
if(now_ans>j-le[i][j]+)ans[j]+=ans[now_ans-];
}
rep(j,,m)ret+=ans[j];
}
printf("%lld\n",ret);
}
//system("Pause");
return ;
}