我只会打尺取法……怎么破

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
using namespace std;
const int MAXN = 1e6 + 5;
int tong[60 + 5];
bool a[MAXN][60 + 5];
int n,k;
vector <int> lsh;
int where(int x)
{
    return lower_bound(lsh.begin(),lsh.end(),x) - lsh.begin();
}
int cqf()
{
    int l = 0,r = -1,minn = 2147483647;
    int ans = 0;
    while(true)
    {
        int maxn = lsh.size();
        while(ans < k && r < maxn)
        {
            r++;
            for(int i = 1;i <= k;i ++)
            {
                if(!a[r][i])
                    continue;
                if(!tong[i])
                    ans++;
                tong[i]++;
            }
        }
        if(ans < k)
            return minn;
        minn = min(minn,lsh[r] - lsh[l]);
        for(int i = 1;i <= k;i ++)
        {
            if(!a[l][i])
                continue;
            tong[i] --;
            if(!tong[i])
                ans--;
        }
        l++;
    }
    return 0;
}
vector <int> z[60 + 5];//x[i] 第i种珠子有这么多个 
int m,x;
int main()
{
    scanf("%d %d",&n,&k);
    for(int i = 1;i <= k;i ++)
    {
        scanf("%d",&m);
        for(int j = 1;j <= m;j ++)
        {
            scanf("%d",&x);
            z[i].push_back(x);
            lsh.push_back(x);
        }
    }
    sort(lsh.begin(),lsh.end());
    unique(lsh.begin(),lsh.end());
    vector<int>::iterator it;
    for(int i = 1;i <= k;i ++)
        for(it = z[i].begin();it != z[i].end();it++)
            a[where(*it)][i] = true;
    printf("%d\n",cqf());
    return 0;
}

看的黄学长博客里的单调队列好神啊……
大家去看吧，我也不敢转载……2333