牛客网Groundhog Looking Dowdy

链接：https://ac.nowcoder.com/acm/contest/5674/F
来源：牛客网

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{n}n days. On the ith{i^{th}}ith day, the jth{j^{th}}jth clothes have a dowdiness ai,ja_{i,j}ai,j​.

On each day,he will choose one of the clothes and wear it.

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

Groundhog wants to know the minimum difference between the maximum dowdiness and the minimum dowdiness in  m{m}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.

翻译：

有n天，每天穿一件衣服，第 i 天有 k_i 件衣服可以穿，穿第 j 件衣服的的权值为 a_{i, j} 。

从 n 天中选择 m 天，求这 m 天中，所穿衣服的权值最大与最小值的最小差是多少。

输入描述:

The first line contains two integers n{n}n and m{m}m.

Then  n{n}n lines follow,each line contains a integer kik_iki​,represents the number of the clothes that can be worn on ith{i^{th}}ith day.Then kik_iki​ integers ai,ja_{i,j}ai,j​ follow.

输出描述:

Print in one line the minimum difference.

1. 由于要最小化最大值和最小值的差值，因此我们可以把所有衣服按照dowdiness从小到大排个序。
2. 排序之后，设最终选出的m件衣服最小覆盖区间为[L,R]，则答案为downdiness[R]-downdiness[L]
3. 则一个合法的区间至少需要包含m种不同的日期
4. 可以对于每个L求出最小的合法的R，这就转化为一个简单的动态窗口问题了。
5. 但是由于数据太水，可以求每天的最小全职并求出最小差即可。

正解代码：

#include
using namespace std;
struct node
{
    int v;
    int h;
}a[2000010];
bool cmp(node x,node y)
{
    return x.v < y.v;
}
int m,cnt=0;
int vis[2000010];
int main()
{
    int n,k;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&k);
        for(int j=1;j<=k;j++)
        {
            cnt++;
            scanf("%d",&a[cnt].v);
            a[cnt].h=i;
        }
    }
    sort(a+1,a+1+cnt,cmp);
    int ans=1e9;
    int l=1,r=m;
    int tmp=0;
    for(int i=l;i<=r;i++)
        if(!vis[a[i].h])
            vis[a[i].h]++;tmp++;
    while(tmp<m)
    {
        r++;
        if(!vis[a[r].h])
        {
            vis[a[r].h]++;
            tmp++;
        }
    }
    for(l,r;r<=cnt;)
    {
        ans=min(ans,a[r].v-a[l].v);
        vis[a[l].h]--;
        l++;
        if(vis[a[l-1].h]==0) tmp--;
        while(tmpcnt)
        {
            r++;
            if(!vis[a[r].h])
            {
                vis[a[r].h]++;
                tmp++;
            }
        }
    }
    printf("%d",ans);
}

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水代码：

#include
using namespace std;
int n,m;
int l,w;
int ans;
int a[1000010];
int main()
{
    for (int i=1; i<=1e6; i++) a[i]=1e9+7;
    ans=1e9+7;
    scanf("%d%d",&n,&m);
    for (int i=1; i<=n; i++)
    {
        scanf("%d",&l);
        for (int j=1; j<=l; j++)
        {
            scanf("%d",&w);
            a[i]=min(w,a[i]);
        }
    }
    sort(a+1,a+1+n);
    for (int i=1; i<=n-m; i++)
        ans=min(ans,a[i+m-1]-a[i]);
    printf("%d",ans);
}

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