[luoguP2564] [SCOI2009]生日礼物（队列）

传送门

不停的枚举 l ，然后枚举长度，直到所有颜色都包含，这是 n² 做法，超时。

仔细想想，可以用个队列来维护。

还是枚举 l ，用队列来维护当前区间的包含所有颜色，l 增加时再判断。

——代码

# include <iostream>
# include <cstdio>
# include <cstring>
# include <string>
# include <cmath>
# include <vector>
# include <map>
# include <queue>
# include <cstdlib>
# include <algorithm>
# define MAXN 1000001
using namespace std;

inline int get_num() {
    int k = 0, f = 1;
    char c = getchar();
    for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for(; isdigit(c); c = getchar()) k = k * 10 + c - '0';
    return k * f;
}

int n, k, tot, sum, h, t, ans = 1 << 30;
int num[MAXN];

struct node
{
    int a, pos;
}p[MAXN];

inline bool cmp(node x, node y)
{
    return x.pos < y.pos;
}

int main()
{
    int i, j, m;
    n = get_num();
    k = get_num();
    for(i = 1; i <= k; i++)
        for(j = 1, m = get_num(); j <= m; j++)
            p[++tot].a = i, p[tot].pos = get_num();
    sort(p + 1, p + n + 1, cmp);
    for(h = 1, t = 0; h <= n; h++)
    {
        while(sum < k && t < n)
        {
            if(!num[p[++t].a]) sum++;
            num[p[t].a]++;
        }
        if(sum == k) ans = min(ans, p[t].pos - p[h].pos);
        num[p[h].a]--;
        if(!num[p[h].a]) sum--;
    }
    printf("%d", ans);
    return 0;
}