Codeforces Round #481 (Div. 3) F. Mentors

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在BerSoft公司，程序员的技能决定了他们能否成为其他程序员的导师。如果程序员a的技能高于程序员b并且他们没有争吵，那么a可以成为b的导师。给定每个程序员的技能和争吵对，找出每个程序员可以成为多少个程序员的导师。解决方案包括排序和动态规划来优化时间复杂度。

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In BerSoft $n$ programmers work, the programmer $i$ is characterized by a skill $r_{i}$ .

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 $b$ $(r_{a} > r_{b})$ 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 $k$ $(2 \leq n \leq 2 \cdot 10^{5}$ , $0 \leq k \leq min (2 \cdot 10^{5}, \frac{n \cdot (n - 1)}{2}))$ — total number of programmers and number of pairs of programmers which are in a quarrel.

The second line contains a sequence of integers $r_{1}, r_{2}, \dots, r_{n}$ $(1 \leq r_{i} \leq 10^{9})$ , where $r_{i}$ equals to the skill of the $i$ -th programmer.

Each of the following $k$ lines contains two distinct integers $x$ , $y$ $(1 \leq x, y \leq n$ , $x \neq y)$ — 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.

题意：一批人求出能做多少个人的师傅，能做师傅的条件是技能k比徒弟高且无争吵

题解: 结构体存一下每个人的序号和技能点，排一下序，求出每个人比他技能点低的人数（n方->nlog n->n）,n方会超时，用dp思想n就行了，然后再通过吵架的--后就是结果了

代码：

#include <iostream>
#include <cmath>
#include <algorithm>
#include <cstring>
using namespace std;

int n,k;
const int M = 2e5+10;
struct Pr
{
    int i, r;
}a[M],b[M];
int ans[M];
int v[M];
int cmp(Pr r1,Pr r2)
{
    return r1.r<r2.r;
}
int main()
{
    memset(ans,0,sizeof(ans));
    memset(v,0,sizeof(v));
    scanf("%d %d",&n,&k);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i].r);
        b[i].r= a[i].r;
        a[i].i = i;
        b[i].i = i;
    }
    sort(a+1,a+1+n,cmp);
    /*超时 
    for(int i=1;i<=n;i++)
    {
        int p=i;
        while(a[p+1].r==a[p].r)
        {
            p++;
        }
        ans[a[i].i] = n-p;
    }
    */
    ans[a[1].i] = 0;
    for(int i=2;i<=n;i++)
    {
    	if(a[i].r!=a[i-1].r)
    		ans[a[i].i] = i-1;
   		else
   			ans[a[i].i] = ans[a[i-1].i];
    }
    /*
    for(int i=1;i<=n;i++)
    {
    	cout<<ans[i]<<" ";
    }
	cout<<endl;
    */
	for(int i=0;i<k;i++)
    {
    	int s1,s2;
    	scanf("%d %d",&s1,&s2);
    	if(b[s1].r>b[s2].r)
   			ans[s1]--;
		if(b[s1].r<b[s2].r)
			ans[s2]--;
    }
    for(int i=1;i<=n;i++)
    {
    	if(ans[i]<0)
    		ans[i] = 0;
        if(i!=n)
            printf("%d ",ans[i]);
        else
            printf("%d\n",ans[i]);
    }
    return 0;
}