Feeling hungry, a cute hamster decides to order some take-away food (like fried chicken for only 3030 Yuan).
However, his owner CXY thinks that take-away food is unhealthy and expensive. So she demands her hamster to fulfill a mission before ordering the take-away food. Then she brings the hamster to a wall.
The wall is covered by square ceramic tiles, which can be regarded as a n * mn∗m grid. CXY wants her hamster to calculate the number of rectangles composed of these tiles.
For example, the following 3 * 33∗3 wall contains 3636 rectangles:
Such problem is quite easy for little hamster to solve, and he quickly manages to get the answer.
Seeing this, the evil girl CXY picks up a brush and paint some tiles into black, claiming that only those rectangles which don't contain any black tiles are valid and the poor hamster should only calculate the number of the valid rectangles. Now the hamster feels the problem is too difficult for him to solve, so he decides to turn to your help. Please help this little hamster solve the problem so that he can enjoy his favorite fried chicken.
Input
There are multiple test cases in the input data.
The first line contains a integer TT : number of test cases. T \le 5T≤5.
For each test case, the first line contains 33 integers n , m , kn,m,k , denoting that the wall is a n \times mn×m grid, and the number of the black tiles is kk.
For the next kk lines, each line contains 22 integers: x\ yx y ,denoting a black tile is on the xx-th row and yy-th column. It's guaranteed that all the positions of the black tiles are distinct.
For all the test cases,
1 \le n \le 10^5,1\le m \le 1001≤n≤105,1≤m≤100,
0 \le k \le 10^5 , 1 \le x \le n, 1 \le y \le m0≤k≤105,1≤x≤n,1≤y≤m.
It's guaranteed that at most 22 test cases satisfy that n \ge 20000n≥20000.
Output
For each test case, print "Case #xx: ansans" (without quotes) in a single line, where xx is the test case number and ansans is the answer for this test case.
Hint
The second test case looks as follows:
样例输入复制
2 3 3 0 3 3 1 2 2
样例输出复制
Case #1: 36 Case #2: 20
题目来源
题意：给出一个矩形，中间有一些挖空的格子，问里面的子矩阵个数。
思路：单调栈维护，对于每一行处理的时候，每次加入一个新的列元素，单调栈里面都更新为height比它小的列，并且从前一个列的的计算结果tmp中减去除去元素的贡献，加入新列元素，再从计算结果tmp中加上他的贡献，最后加到ans中去。时间复杂度O(n*m)。
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5+10;
const int maxm = 1e2+10;
bool box[maxn][maxm];
int up[maxm];
int ht[maxm];
int id[maxm];
int main()
{
int t;
int n,m,k;
scanf("%d",&t);
int tt = 0;
while(t--)
{
memset(box,0,sizeof(box));
memset(up,0,sizeof(up));
scanf("%d%d%d",&n,&m,&k);
int x,y;
for(int i = 0;i<k;i++)
{
scanf("%d%d",&x,&y);
box[x][y] = 1;
}
ll ans = 0;
for(int i = 1;i<=n;i++)
{
ll tmp = 0;
ll sum = 0;
int cnt = 0;
for(int j = 1;j<=m;j++)
{
if(box[i][j])
{
up[j] = i;
tmp = 0;
ht[0] = 0;
id[0] = j;
cnt = 1;
continue;
}
int h = i - up[j];
while(cnt>0&&h<ht[cnt-1])
{
if(cnt - 1>0)
{
tmp -= (id[cnt-1] - id[cnt-2])*ht[cnt-1];
}
else
{
tmp -= ht[cnt-1]*(id[cnt-1]);
}
cnt--;
}
if(cnt)
{
tmp += (j - id[cnt-1])*h;
}
else
{
tmp += j*h;
}
ht[cnt] = h;
id[cnt++] = j;
sum += tmp;
}
ans += sum;
}
printf("Case #%d: %lld\n",++tt,ans);
}
}