Codeforces Round #443 (Div. 1) C. Tournament

C. Tournament

link

http://codeforces.com/contest/878/problem/C

description

Recently a tournament in k kinds of sports has begun in Berland. Vasya wants to make money on the bets.

The scheme of the tournament is very mysterious and not fully disclosed. Competitions are held back to back, each of them involves two sportsmen who have not left the tournament yet. Each match can be held in any of the k kinds of sport. Loser leaves the tournament. The last remaining sportsman becomes the winner. Apart of this, the scheme can be arbitrary, it is not disclosed in advance.

Vasya knows powers of sportsmen in each kind of sport. He believes that the sportsmen with higher power always wins.

The tournament is held every year, and each year one new participant joins it. In the first tournament, only one sportsman has participated, in the second there were two sportsmen, and so on. Vasya has been watching the tournament for the last n years. Help him to find the number of possible winners for each of the n tournaments.

Input

The first line contains two integers n and k (1 ≤ n ≤ 5·104, 1 ≤ k ≤ 10) — the number of tournaments and the number of kinds of sport, respectively.

Each of the next n lines contains k integers si1, si2, ..., sik (1 ≤ sij ≤ 109), where sij is the power of the i-th sportsman in the j-th kind of sport. The sportsman with higher powers always wins. It's guaranteed that for any kind of sport all of these powers are distinct.

Output

For each of the n tournaments output the number of contenders who can win.

Examples

input

3 2
1 5
5 1
10 10

output

1
2
1

input

3 2
2 2
3 3
1 10

output

1
1
3

input

3 2
2 3
1 1
3 2

output

1
1
2

Note

In the first sample:

In the first tournament there is only one sportsman, and he is the winner.

In the second tournament, there are two sportsmen, and everyone can defeat another, depending on kind of sports.

In the third tournament, the third sportsman in the strongest in both kinds of sports, so he is the winner regardless of the scheme.

题意

如今有n我的参加比赛，每一个人都有k项能力值，每一个能力值表示这我的在这个技能上面的能力。
如今这些人会两两进行比赛，他们会挑选一个技能进行比较，分值高的获胜，胜利者留下来，失败者离开。
你须要回答，最后究竟会有多少我的可能成为冠军。

题解

假设如今有一群人均可能得到冠军，那么必须知足任何一我的，都能找到一我的的某项技能小于等于他，也能找到一我的的某项技能强于他。
那么对于这个集合，对于每一个人都能存在一个比赛方案，使得他成为冠军。
而后咱们动态维护这个集群就行了，用set去维护。

你能够用图论的方式去理解。若是A可以战胜B，那么连一条A->B的边，显然胜利者的含义就是若是A可以达到其余的全部点，那么A就是胜利者。
最后的胜利者集合，里面的任何两个点都能互相到达，若是已经成为了团，咱们就进行缩点就行了，咱们用set去缩点。
而后维护每个团就行。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+7;
int n,k;
struct node{
    int mx[10],mi[10],sz;
    bool operator < (const node &b)const{
        for(int i=0;i<k;i++){
            if(mx[i]>b.mi[i])return false;
        }
        return true;
    }
};
set<node>S;
int main(){
    scanf("%d%d",&n,&k);
    for(int i=0;i<n;i++){
        node a;
        for(int j=0;j<k;j++){
            scanf("%d",&a.mx[j]);
            a.mi[j]=a.mx[j];a.sz=1;
        }
        set<node>::iterator x = S.lower_bound(a);
        set<node>::iterator y = S.upper_bound(a);
        while(x!=y){
            a.sz+=x->sz;
            for(int j=0;j<k;j++){
                a.mi[j]=min(a.mi[j],x->mi[j]);
                a.mx[j]=max(a.mx[j],x->mx[j]);
            }
            S.erase(x++);
        }
        S.insert(a);
        cout<<S.rbegin()->sz<<endl;
    }
}