Codeforces Round #481 (Div. 3) F. Mentors（利用结构体）简单方便

F. Mentors

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 riri equals 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

Copy

4 2
10 4 10 15
1 2
4 3

output

Copy

0 0 1 2 

input

Copy

10 4
5 4 1 5 4 3 7 1 2 5
4 6
2 1
10 8
3 5

output

Copy

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.

 

大意：如果一个程序员比另一个程序员的能力值高，而且这两个程序员不在争吵状态，则能力值较高的程序员可以成为另一个程序员的老师，求每个程序员能成为多少其他程序员的老师。

思路：可以先不管争吵状态，先求出每个程序员可以当多少人的老师。然后再把争吵状态中的：能力值大的程序员（且可以当别人的老师（即b[i]!=0））当别人老师数减一（即b[i]-1）.该题用暴搜会超时，所以我利用了结构体和dp思想：先把程序员能力和编号读再同一结构体a【MAXN】中，然后排序，当a[i].r!=a[i-1].r（说明该程序员能力值大于前面所有程序员的能力值）时刚好这个程序员可当老师的人数为i-1；而a[i].r==a[i-1].r时，该程序员与前面程序员的能力值相等，当别人老师数也就相等（现在不考虑争吵状态）即[a[i].q]=b[a[i-1].q].复杂度为sort函数复杂度 nlogn.

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
#define MAXN 210000
struct P{int q,r;}a[MAXN];
int b[MAXN],s[MAXN];
int cmp(P a,P b)
{
return a.r<b.r;
}
int main()
{int n,m,i,j,c,d;
while(scanf("%d%d",&n,&m)!=EOF)
{
  for(i=1;i<=n;i++)
  {scanf("%d",&a[i].r);a[i].q=i;
  s[i]=a[i].r;}
  sort(a+1,a+n+1,cmp);
  for(i=2;i<=n;i++)
if(a[i].r!=a[i-1].r)
         b[a[i].q]=i-1;
else b[a[i].q]=b[a[i-1].q];
   for(i=1;i<=m;i++)
  {int x,y;
  scanf("%d%d",&x,&y);
  if(s[x]<s[y]&&b[y]!=0)
 b[y]--;
  if(s[x]>s[y]&&b[x]!=0)
 b[x]--;
  }
  printf("%d",b[1]);
  for(i=2;i<=n;i++)
 printf(" %d",b[i]);
  printf("\n");
}
return 0;
}