codeforces+contest978F. Mentors+思维

F. Mentors
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

In BerSoft n

programmers work, the programmer i is characterized by a skill ri

.

A programmer a

can be a mentor of a programmer b if and only if the skill of the programmer a is strictly greater than the skill of the programmer b (ra>rb) and programmers a and b

are not in a quarrel.

You are given the skills of each programmers and a list of k

pairs of the programmers, which are in a quarrel (pairs are unordered). For each programmer i, find the number of programmers, for which the programmer i

can be a mentor.

Input

The first line contains two integers n

and k (2n2105, 0kmin(2105,n(n1)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,,rn

(1ri109), where ri equals to the skill of the i

-th programmer.

Each of the following k

lines contains two distinct integers x, y (1x,yn, xy) — pair of programmers in a quarrel. The pairs are unordered, it means that if x is in a quarrel with y then y is in a quarrel with x. Guaranteed, that for each pair (x,y) there are no other pairs (x,y) and (y,x)

in the input.

Output

Print n

integers, the i-th number should be equal to the number of programmers, for which the i

-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.

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define all(a) (a).begin(),(a).end()
#define pll pair<ll,ll>
#define vi vector<int>
#define pb push_back
ll rd(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

int main(){
   freopen("in.txt","r",stdin);
  int n=rd(),m=rd();
  vector<pll> a(n);
  vector<ll> init(n);
  rep(i,0,n-1){
      a[i].first=rd();
      a[i].second=i;
      init[i]=a[i].first;
  }
  sort(all(a));
  vector<int> ans(n);
  vector<vi> g(n);
  while(m--){
    int u=rd()-1,v=rd()-1;
    g[u].pb(v);
    g[v].pb(u);
  }
  int cnt=0;
  int curr=0;
  rep(i,0,n-1){
      ++curr;
      int z=cnt;
      for(auto to:g[a[i].second])
      if(init[a[i].second]>init[to])
        z--;
        ans[a[i].second]=z;
      if(i+1<n&&a[i].first!=a[i+1].first){
        cnt+=curr;
        curr=0;
      }
  }
  rep(i,0,n-1)
  printf("%d ",ans[i]);
  return puts(""),0;
}

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转载自blog.csdn.net/ujn20161222/article/details/80308505