Codeforces 651C Watchmen【模拟】

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……..