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Codeforces Round #345 (Div. 2) C

2019-11-14 12:04:57
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C. Watchmen time limit per test3 seconds memory limit per test256 megabytes inputstandard input outputstandard output Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are n watchmen on a plane, the i-th watchman is located at point (xi,?yi).

They need to arrange a plan, but there are some difficulties on their way. As you know, Doctor Manhattan considers the distance between watchmen i and j to be |xi?-?xj|?+?|yi?-?yj|. Daniel, as an ordinary person, calculates the distance using the formula .

The success of the Operation relies on the number of pairs (i,?j) (1?≤?i?<?j?≤?n), such that the distance between watchman i and watchmen j calculated by Doctor Manhattan is equal to the distance between them calculated by Daniel. You were asked to compute the number of such pairs.

Input The first line of the input contains the single integer n (1?≤?n?≤?200?000) — the number of watchmen.

Each of the following n lines contains two integers xi and yi (|xi|,?|yi|?≤?109).

Some positions may coincide.

Output PRint the number of pairs of watchmen such that the distance between them calculated by Doctor Manhattan is equal to the distance calculated by Daniel.

Examples input 3 1 1 7 5 1 5 output 2 input 6 0 0 0 1 0 2 -1 1 0 1 1 1 output 11 Note In the first sample, the distance between watchman 1 and watchman 2 is equal to |1?-?7|?+?|1?-?5|?=?10 for Doctor Manhattan and for Daniel. For pairs (1,?1), (1,?5) and (7,?5), (1,?5) Doctor Manhattan and Daniel will calculate the same distances.

求曼哈頓距離和距離相同的對數… 就是重復x+重復y-重復xy 但是這個hash過程要記住..

#include<iostream>#include<algorithm>#include<map>#include<vector>using namespace std;long long tx[200001],txs[200001] ,tys[200001],ty[200001],hsx[200001],hsy[200001],hsxy[200001];map<long long,long long>x, y,xy[200001];vector<long long>dq[200001];long long suan(long long q){ q--; return (q + 1)*q / 2;}int main(){#define int long long int n; cin >> n; for (int a = 1;a <= n;a++)scanf("%I64d%I64d", &tx[a], &ty[a]); for (int a = 1;a <= n;a++) { if (!x[tx[a]])x[tx[a]] = x.size(); if (!y[ty[a]])y[ty[a]] = y.size(); int xq = x[tx[a]], yq = y[ty[a]]; txs[xq]++, tys[yq]++; xy[xq][yq]++; if (xy[xq][yq] == 2)dq[xq].push_back(yq); } int js = 0; for (int a = 1;a <= n;a++) { int xq = x[tx[a]]; if (hsx[xq])continue; js += suan(txs[xq]); hsx[xq] = 1; } for (int a = 1;a <= n;a++) { int yq = y[ty[a]]; if (hsy[yq])continue; js += suan(tys[yq]); hsy[yq] = 1; } for (int a = 1;a <= n;a++) { int xq = x[tx[a]]; if (hsxy[xq])continue; hsxy[xq] = 1; for (int a = 0;a < dq[xq].size();a++)js -= suan(xy[xq][dq[xq][a]]); } cout << js;}
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