2016 ACM Amman Collegiate Programming Contest D Rectangles

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

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

output

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\n",ret);
    }
    //system("Pause");
    return 0;
}