2020牛客暑期多校训练营（第九场）F.Groundhog Looking Dowdy

2020牛客暑期多校训练营（第九场）F.Groundhog Looking Dowdy

题目链接

题目描述

Groundhog finds that Apple seems to be less intimate than before.

He is very distressed for this.After pondering for a long time, Groundhog finds that he looks too dowdy.So, Groundhog decided to improve a little.

Because Groundhog is lazy, he only lists the clothes that can be worn for the next $n$ days. On the $i^{th}$ day, the $j^{th}$ clothes have a dowdiness $a_{i,j}$.On each day,he will choose one of the clothes and wear it.

And Groundhog should choose $m$ days from $n$ days to go out with Apple.

Groundhog wants to know the minimum difference between the maximum dowdiness and the minimum dowdiness in $m$ days when Groundhog’s choice is optimal.

Simplified: You should choose n clothes for each day and then choose m clothes from those n clothes, and the problem is to calculate the minimum difference between the maximum dowdiness and the minimum dowdiness.

输入描述:

The first line contains two integers $n$ and $m$.
Then {n}n lines follow,each line contains a integer $k_i$,represents the number of the clothes that can be worn on $i^{th}$ day.Then $k_i$ integers $a_{i,j}$ follow.

输出描述:

Print in one line the minimum difference.

示例1

输入

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

输出

2

典型的尺取，存两个值，一个 $dowdiness$，还有一个下标 $i\in [1,n]$，按 $dowdiness$ 进行排序，然后尺取即可，每当种类达到 $m$ 就更新答案即可，尺取的难点就在于左端点的更新，注意细节即可。吐槽一下，我看有通过代码有人直接按最小值排序混过的，很无语，证明数据还是太水了，AC代码如下：

#include
using namespace std;
const int N=2e6+5;
vector<pair<int,int>>good;
int mp[N]={0};
int main()
{
    int n,m,k,x;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        scanf("%d",&k);
        for(int j=0;j<k;j++){
            scanf("%d",&x);
            good.push_back({x,i});
        }
    }
    sort(good.begin(),good.end());
    int l=0,cnt=0,ans=1e9,len=good.size();
    for(int i=0;i<len;i++){
        if(mp[good[i].second]==0){
            mp[good[i].second]++;
            cnt++;
        }else mp[good[i].second]++;
        while(cnt>=m){
            if(mp[good[l].second]>1) mp[good[l].second]--,l++;
            else{
                if(cnt-1>=m) mp[good[l].second]=0,l++,cnt--;
                else break;
            }
        }
        if(cnt==m) ans=min(ans,good[i].first-good[l].first);
    }
    printf("%d",ans);
	return 0;
}