codeforces #345 div 2 C. Watchmen (容斥)

题目链接 题意：求曼哈顿距离和平方根距离相等的点的对数？ 思路：化简发现是绝对值乘积等于0，容斥搞搞。

/* ***********************************************
Author :111qqz
Created Time :2016年03月07日 星期一 18时43分02秒
File Name :code/C.cpp
************************************************ */

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair

using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const int N=2E6+7;

struct node
{
    int x,y;

    bool operator < (node b)const
    {
	if (x==b.x) return y<b.y;
	return x<b.x;
    }
}p[N];



bool cmp( node a,node b)
{
    return a.y<b.y;
}
LL cal( LL x)
{
    LL res = x*(x+1)/2;
    return res;
}
int n;
LL a,b,c;
LL cnta,cntb,cntc;
int main()
{
	#ifndef  ONLINE_JUDGE 
	freopen("code/in.txt","r",stdin);
  #endif

	ios::sync_with_stdio(false);
	cin>>n;
	for ( int i = 1 ;i  <= n ; i++) cin>>p[i].x>>p[i].y;

	sort(p+1,p+n+1);

	a=b=c=0LL;
	cnta=cntb=cntc=0LL;


	for ( int i = 1 ; i <= n-1 ; i++)
	{
	    if (p[i].x==p[i+1].x)
	    {
		cnta++;
	    }
	    else
	    {
		a+=cal(cnta);
		cnta = 0 ;
	    }
	}
	a +=cal(cnta);


	for ( int i = 1 ; i <= n-1 ; i++)
	{
	    if (p[i].x==p[i+1].x&&p[i].y==p[i+1].y)
	    {
		cntc++;
	    }
	    else
	    {
		c +=cal(cntc);
		cntc = 0 ;
	    }
	}

	c+=cal(cntc);



	sort(p+1,p+n+1,cmp);
	for ( int i =  1 ; i  <= n-1 ; i ++)
	{
	    if (p[i].y==p[i+1].y)
	    {
		cntb++;
	    }
	    else
	    {
		b +=cal(cntb);
		cntb = 0 ;
	    }
	}
	b +=cal(cntb);

	LL ans=0LL;
	ans = a+b-c;
	cout<<ans<<endl;

  #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}