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.
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.
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.
3
1 1
7 5
1 5
2
6
0 0
0 1
0 2
-1 1
0 1
1 1
11
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.
题目是让你求有多少点，它们的曼哈顿距离 等于 欧几里得距离。就是(xi - xj) * (yi - yj) == 0;
用到了容斥原理，计算 xi & xj 相等的点有多少个，计算 yi & yj 相等的点有多少个，然后再减去 xi & xj 和 yi & yj 都想的点有多少个。
package codefroces345;
import java.io.*;
import java.util.*;
public class C345{
/*
* java io 系统给的这么慢。。。时间是优化后的3倍。。。
* */
static class Pair{
int x, y;
public Pair(int x, int y){
this.x = x;
this.y = y;
}
@Override
public int hashCode(){
final int prime = 31;
int result = 1;
result = prime * result + x;
result = prime * result + y;
return result;
}
@Override
public boolean equals(Object obj){
if(this == obj){
return true;
}
if(this == null) {
return false;
}
if(getClass() != obj.getClass()){
return false;
}
Pair other = (Pair)obj;
if(x != other.x)
return false;
if(y != other.y)
return false;
return true;
}
}
public static void main(String[] args){
Scanner scanner = new Scanner(new InputStreamReader(System.in));
int n;
n = scanner.nextInt();
HashMap<Integer, Integer> xs = new HashMap<>();
HashMap<Integer, Integer> ys = new HashMap<>();
HashMap<Pair, Integer> both = new HashMap<>();
for(int i = 0; i < n; ++i) {
int x = scanner.nextInt();
int y = scanner.nextInt();
Pair p = new Pair(x, y);
xs.put(x, xs.getOrDefault(x, 0) + 1);
ys.put(y, ys.getOrDefault(y, 0) + 1);
both.put(p, both.getOrDefault(p, 0) + 1);
}
long ans = 0;
for(int v : xs.values()){
ans += (long) v * (v-1) / 2;
}
for(int v : ys.values()){
ans += (long) v * (v-1) / 2;
}
for(int v : both.values()) {
ans -= (long) v * (v-1) / 2;
}
System.out.println(ans);
}
}
看人家们的代码，用到了BufferedReader包装加速
package codefroces345;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.HashMap;
import java.util.StringTokenizer;
/**
* Created by lenovo on 2016-03-10.
*/
/*
* 经过 i/o 包装后，只要600ms左右，还是学的少
* */
public class C {
BufferedReader br;
PrintWriter out;
StringTokenizer st;
boolean eof;
static class Pair {
int x, y;
public Pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + x;
result = prime * result + y;
return result;
}
@Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
Pair other = (Pair) obj;
if (x != other.x)
return false;
if (y != other.y)
return false;
return true;
}
}
void solve() throws IOException {
int n = nextInt();
HashMap<Integer, Integer> xs = new HashMap<>();
HashMap<Integer, Integer> ys = new HashMap<>();
HashMap<Pair, Integer> both = new HashMap<>();
for (int i = 0; i < n; i++) {
int x = nextInt();
int y = nextInt();
Pair p = new Pair(x, y);
xs.put(x, xs.getOrDefault(x, 0) + 1);
ys.put(y, ys.getOrDefault(y, 0) + 1);
both.put(p, both.getOrDefault(p, 0) + 1);
}
long ans = 0;
for (int v : xs.values()) {
ans += (long)v * (v - 1) / 2;
}
for (int v : ys.values()) {
ans += (long)v * (v - 1) / 2;
}
for (int v : both.values()) {
ans -= (long)v * (v - 1) / 2;
}
out.println(ans);
}
C() throws IOException{
br = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
solve();
out.close();
br.close();
}
public static void main(String[] args) throws IOException{
new C();
}
String nextToken(){
while(st == null || !st.hasMoreTokens()){
try{
st = new StringTokenizer(br.readLine());
} catch(IOException e) {
eof = true;
return null;
}
}
return st.nextToken();
}
String nextString(){
try{
return br.readLine();
} catch (Exception e) {
eof = true;
return null;
}
}
int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}
long nextLong() throws IOException {
return Long.parseLong(nextToken());
}
double nextDouble() throws IOException {
return Double.parseDouble(nextToken());
}
}