Mentors CodeForces - 978F （思维 stl的应用）

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 $r_i$.

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

The second line contains a sequence of integers $r_1,r_2,\dots ,r_n$ $(1\le r_i\le {10}^9)$, where $r_i$ equals to the skill of the $i$-th programmer.

Each of the following $k$ lines contains two distinct integers $x$, $y$ $(1\le x,y\le n$, $x\ne y)$ — 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：给出n个人的能力值，给出m对关系，如果一个人的能力值比另一个人的能力值高，并且他们两没有关系的话，那么，这个人就可以成为那个人的导师。求出每个人最多可以是几个人的导师。

思路：结构体来记录下标，能力值，和结果。然后按能力值升序排序一下，用排序后的下标减去有关系的人的个数就是答案了。在这里，我们只记录从能力值高的到能力值低的人之间的关系。反过来不记录。如果反过来也记录的话，就会多减去一些值！注意，这里如果有能力值相同的，也会被计算，所以用一个lower_bound来找出当前节点之前与之能力值相同的点的个数，并将之减去。（在这里，我不会在结构体里面用lower_bound查找，所以重新开了一个一模一样的数组来查找  真丢人呐！！）最后，按下标排序，输出即可。

#include "iostream"
#include "algorithm"
#include "vector"
using namespace std;
const int Max=2e5+10;
struct pot
{
    int date,index;
    int ans;
};
pot tmp[Max];
bool cmp1(pot a,pot b)
{
    return a.date<b.date;
}
bool cmp2(pot a,pot b)
{
    return a.index<b.index;
}
vector<int> G[Max];
int a[Max];
int main()
{
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        cin>>tmp[i].date;
        a[i]=tmp[i].date;
        tmp[i].index=i;
        tmp[i].ans=0;
        G[i].clear();
    }
    int u,v;
    for(int i=1;i<=m;i++){
        cin>>u>>v;
        if(tmp[u].date>tmp[v].date)//只记录能力值从高到低的关系
            G[u].push_back(v);
        else if(tmp[u].date<tmp[v].date)
            G[v].push_back(u);
    }
    sort(tmp+1,tmp+1+n,cmp1);
    sort(a+1,a+1+n);
    for(int i=2;i<=n;i++){
        int x=a[i];
        if(x==a[1]) continue;
        int k=i-(lower_bound(a+1,a+1+n,x)-a);//找出之前，与之能力值相通的点的个数
        int y=i-G[tmp[i].index].size()-1-k;
        tmp[i].ans=y;
    }
    sort(tmp+1,tmp+1+n,cmp2);
    for(int i=1;i<=n;i++){
        cout<<tmp[i].ans;
        if(i!=n)
            cout<<" ";
    }
    cout<<endl;
    return 0;
}

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