STL应用---Codeforces-650-A.watchman

A. Watchmen
time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard 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.

题目大意：求出曼哈顿距离与欧几里得距离相同的总的对数；
解题思路：根据公式：
与公式：|xi - xj| + |yi - yj|，可以将其同时平方，然后化简，得出二者相等的条件是:xi==xj || yi==yj .
然后可以 STL的映射map来求解，方便快捷；

具体代码如下：

#include <cstdio>#include<algorithm>#include <map>using namespace std;int n,x,y;long long m;map<int,int> a,b;map<pair<int,int>,int> c;int main(){    scanf("%d",&n);    while (n--)    {        scanf("%d%d",&x,&y);        m+=a[x]+b[y]-c[make_pair(x,y)];//注意减去重复的        ++a[x];++b[y];        ++c[make_pair(x,y)];    }        printf("%I64d\n",m);        return 0; }

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