F - Mentors
In BerSoft nn programmers work, the programmer ii is characterized by a skill riri.
A programmer aa can be a mentor of a programmer bb if and only if the skill of the programmer aa is strictly greater than the skill of the programmer bb (ra>rb)(ra>rb) and programmers aa and bb are not in a quarrel.
You are given the skills of each programmers and a list of kk pairs of the programmers, which are in a quarrel (pairs are unordered). For each programmer ii, find the number of programmers, for which the programmer ii can be a mentor.
Input
The first line contains two integers nn and kk (2≤n≤2⋅105(2≤n≤2⋅105, 0≤k≤min(2⋅105,n⋅(n−1)2))0≤k≤min(2⋅105,n⋅(n−1)2)) — total number of programmers and number of pairs of programmers which are in a quarrel.
The second line contains a sequence of integers r1,r2,…,rnr1,r2,…,rn (1≤ri≤109)(1≤ri≤109), where ririequals to the skill of the ii-th programmer.
Each of the following kk lines contains two distinct integers xx, yy (1≤x,y≤n(1≤x,y≤n, x≠y)x≠y) — pair of programmers in a quarrel. The pairs are unordered, it means that if xx is in a quarrel with yy then yy is in a quarrel with xx. Guaranteed, that for each pair (x,y)(x,y) there are no other pairs (x,y)(x,y) and (y,x)(y,x) in the input.
Output
Print nn integers, the ii-th number should be equal to the number of programmers, for which the ii-th programmer can be a mentor. Programmers are numbered in the same order that their skills are given in the input.
Examples
Input
4 2
10 4 10 15
1 2
4 3
Output
0 0 1 2
Input
10 4
5 4 1 5 4 3 7 1 2 5
4 6
2 1
10 8
3 5
Output
5 4 0 5 3 3 9 0 2 5
Note
In the first example, the first programmer can not be mentor of any other (because only the second programmer has a skill, lower than first programmer skill, but they are in a quarrel). The second programmer can not be mentor of any other programmer, because his skill is minimal among others. The third programmer can be a mentor of the second programmer. The fourth programmer can be a mentor of the first and of the second programmers. He can not be a mentor of the third programmer, because they are in a quarrel.
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstring>
using namespace std;
const int MAX=2e5+7;
int weizhi(int* arr,int r,int n) //返回第i个人的能力排行名次,arr位已经排序好的能力数组,r为第i个人的能力值
{
return lower_bound(arr,arr+n,r)-arr; //lower_bound()函数 返回大于或等于r的位置,我的博客中一片篇有详细的讲解
}
int main()
{
int n,m;
int arr[MAX]; //原数组
int brr[MAX]; //能力排序数组
int ans[MAX]; //标记数组
cin >>n>>m;
for(int i=0;i<n;i++)
{
cin >>arr[i];
brr[i]=arr[i];
}
memset(ans,0,sizeof(ans));
sort (brr,brr+n); //能力排序
for(int i=0;i<m;i++)
{
int a,b;
cin >>a>>b;
a--;
b--; //原数组下标是从零开始的,因此要减一
if(arr[a]>arr[b])
ans[a]--; //能力大的要排在能力小的后边,如果两个人吵架的话,只对能力大的人有影响(能力排在前边的的不能当成为能力大的人的学生)
else if (arr[a]<arr[b]) //此处要写小于,如果写,当两个人能力值相等并且吵架的话,ans就会多减
ans[b]--;
}
for(int i=0;i<n;i++)
{
int res=weizhi(brr,arr[i],n)+ans[i]; //输出可以当几个人的老师
cout <<res<<" ";
}
}