题意：

求欧几里得距离与曼哈顿距离相等的组数。

分析：

化简后得到xi=xj||yi=yj，即为求x相等 + y相等 - x与y均相等。

代码：

#include<iostream>

#include<algorithm>

using namespace std;

const int maxn = 1000000 + 5;

typedef pair<long long , long long>pii;

pii p[maxn];

bool cmp(pii a, pii b)

{

    return a.first < b.first||(a.first == b.first && a.second < b.second);

}

bool cmp1(pii a, pii b)

{

    return a.second < b.second||(a.second == b.second &&a.first < b.first);

}

int main (void)

{

    int n;cin>>n;

    for(int i = 0; i < n; i++){

       cin>>p[i].first>>p[i].second;

    }

    long long resx = 0, resy = 0;

    long long cnt = 0;

    sort(p, p + n, cmp);

    for(int i = 1; i < n; i++){

       if(p[i].first == p[i - 1].first)

            cnt++;

        if(i == n-1||p[i].first != p[i - 1].first){

           resx += cnt * (cnt + 1)/2;

            cnt = 0;

        }

    }

    sort(p, p + n, cmp1);

    cnt = 0;

    long long  aa = 0, both = 0;

    for(int i = 1; i < n; i++){

        if(p[i].second == p[i - 1].second){

            cnt++;

            if(p[i].first == p[i-1].first)

                aa++;

            if(p[i].first != p[i-1].first){

                both += aa * (aa + 1)/2;

                aa = 0;

            }

        }

        if(i == n-1||p[i].second != p[i - 1].second){

            resy += cnt * (cnt + 1)/2;

            cnt = 0;

            both += aa * (aa + 1)/2;

            aa = 0;

        }

    }

    cout<<resx + resy - both<<endl;

}

也可以用map做，这样可以直接记录x和y均相等的个数。

#include <iostream>

#include <cstring>

#include <algorithm>

#include <map>

using namespace std;

const int maxn = 200005;

typedef pair<int, int> p;

typedef long long ll;

map<int, int>ma, mb;

map<p, int> mp;

int main (void)

{

    int n;cin>>n;

    int x, y;

    for(int i = 0; i < n; i++){

        cin>>x>>y;

        ma[x] ++;

        mb[y]++;

        mp[p(x, y)]++;

    }

    long long resx = 0, resy = 0, both = 0;

    map<int, int>::iterator i;

    map<p, int>::iterator j;

    for(i = ma.begin(); i != ma.end(); i++){

        resx +=(ll) i->second * ( i->second- 1)/2;

    }

    for(i = mb.begin(); i!=mb.end();i++){

        resy += (ll)i->second * ( i->second- 1)/2;

    }

    for(j = mp.begin(); j != mp.end(); j++){

        both += (ll) j->second * ( j->second- 1)/2;

    }

    cout<<resx + resy - both<<endl;

    return 0;

}

对于C++11，for(i = ma.begin(); i != ma.end(); i++)可以直接写成

for(auto x: ma) 

不明白第一种方法如果不分别保存resx,resy,both而是直接对res进行不停的加减，最后会runtime error……..