Problem Description
Matt loves letter L.
A point set P is (a, b)-L if and only if there exists x, y satisfying:
P = {(x, y), (x + 1, y), . . . , (x + a, y), (x, y + 1), . . . , (x, y + b)}(a, b ≥ 1)
A point set Q is good if and only if Q is an (a, b)-L set and gcd(a, b) = 1.
Matt is given a point set S. Please help him find the number of ordered pairs of sets (A, B) such that:
Input
The first line contains only one integer T , which indicates the number of test cases.
For each test case, the first line contains an integer N (0 ≤ N ≤ 40000), indicating the size of the point set S.
Each of the following N lines contains two integers xi, yi, indicating the i-th point in S (1 ≤ xi, yi ≤ 200). It’s guaranteed that all (xi, yi) would be distinct.
Output
For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1) and y is the number of pairs.
Sample Input
2
6
1 1
1 2
2 1
3 3
3 4
4 3
9
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
Sample Output
Case #1: 2
Case #2: 6
Hint
n the second sample, the ordered pairs of sets Matt can choose are:
A = {(1, 1), (1, 2), (1, 3), (2, 1)} and B = {(2, 2), (2, 3), (3, 2)}
A = {(2, 2), (2, 3), (3, 2)} and B = {(1, 1), (1, 2), (1, 3), (2, 1)}
A = {(1, 1), (1, 2), (2, 1), (3, 1)} and B = {(2, 2), (2, 3), (3, 2)}
A = {(2, 2), (2, 3), (3, 2)} and B = {(1, 1), (1, 2), (2, 1), (3, 1)}
A = {(1, 1), (1, 2), (2, 1)} and B = {(2, 2), (2, 3), (3, 2)}
A = {(2, 2), (2, 3), (3, 2)} and B = {(1, 1), (1, 2), (2, 1)}
Hence, the answer is 6.
题意：对于点集P 如果存在a,b使得P = {(x, y), (x + 1, y), . . . , (x + a, y), (x, y + 1), . . . , (x, y + b)}(a, b ≥ 1)，并且a，b互质，则P is good 。可以发现对于符合要求（good）的集合P ，其构成一个L 型，且以（x,y）为拐点，从（x,y）向上长度和向右长度互质。现在给了N个点，求有多少对符合要求的L型集合不相交（集合交集为空）？
思路：先找到所有符合要求的L个数S，那么用S*S-相交的L对数 即为结果。
怎么算相交的所有L对数呢？ 容斥，很妙的思想，遍历每一个点，如果当前的点是输入的点之一，那么是一个拐点，令这个拐点向右延伸最长为k，那么算出所有其它L的竖着部分与（x,y+k）相交的对数，乘以2，另外要考虑以（x,y）为拐点的L与自身相交的情况，把这两种相交情形减掉后既是结果。
代码如下：
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;
const int N=40050;
const int M=205;
int R[M][M],U[M][M];
bool mp[M][M];
int dp[M][M],cnt[M][M];
int t[M][M];
int gcd(int a,int b) { return (b==0)?a:gcd(b,a%b); }
void init()
{
for(int i=1;i<M;i++)
for(int j=1;j<M;j++)
{
dp[i][j]=dp[i][j-1]+((gcd(i,j)==1)?1:0);
cnt[i][j]=cnt[i-1][j]+dp[i][j];
}
}
int main()
{
init();
int T,Case=1;
cin>>T;
while(T--)
{
int n; scanf("%d",&n);
memset(mp,0,sizeof(mp));
memset(U,0,sizeof(U));
memset(R,0,sizeof(R));
memset(t,0,sizeof(t));
for(int i=1;i<=n;i++)
{
int x,y;
scanf("%d%d",&x,&y);
mp[x][y]=1;
}
for(int i=200;i>=1;i--)
{
for(int j=200;j>=1;j--)
{
if(mp[i][j]){
if(mp[i+1][j]) U[i][j]=U[i+1][j]+1;
if(mp[i][j+1]) R[i][j]=R[i][j+1]+1;
}
}
}
LL s=0;
for(int i=1;i<=200;i++)
{
for(int j=1;j<=200;j++)
{
if(mp[i][j]){
s+=cnt[U[i][j]][R[i][j]];
int d=0;
for(int k=U[i][j];k>=0;k--)
{
d+=dp[k][R[i][j]];
t[i+k][j]+=d;
}
}
}
}
LL ans=0;
for(int i=1;i<=200;i++)
{
for(int j=1;j<=200;j++)
{
if(mp[i][j]){
LL p=t[i][j];
LL pp=cnt[U[i][j]][R[i][j]];
p-=pp;
for(int k=1;k<=R[i][j];k++)
{
p+=t[i][j+k];
ans+=2*p*dp[k][U[i][j]];
}
ans+=pp*pp;
}
}
}
s=s*s-ans;
printf("Case #%d: %lld\n",Case++,s);
}
return 0;
}