[SCOI2009]生日礼物

思路

如果没有记错的话,这题跟某次多校的题几乎一模一样。
区间问题的最小值,有一个最简单的办法就是尺取法了,只要通过两个指针的扫描就可以以线性的复杂度简单的实现,这里我们按照每个物品属于的种类作为我们num数组记录的值,然后再通过一次扫描就可以得到答案了。

代码

/*
  Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>

#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;

inline ll read() {
    ll f = 1, x = 0;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-')    f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return f * x;
}

const int N = 1e6 + 10;

struct Node {
    int pos, id;
    bool operator < (const Node & t) const {
        return pos < t.pos;
    }
}a[N];

int num[N], n, k;

int main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    n = read(), k = read();
    int tot = 0;
    for(int i = 1; i <= k; i++) {
        int x = read();
        for(int j = 1; j <= x; j++) {
            int t = read();
            tot++;
            a[tot].id = i;
            a[tot].pos = t;
        }
    }
    sort(a + 1, a + 1 + tot);
    int ans = inf, l = 1, r = 1, cnt = 0;
    for(;;) {
        while(r <= tot && cnt < k) {
            cnt += !num[a[r++].id]++;
        }
        if(cnt < k) break;
        ans = min(ans, a[r - 1].pos - a[l].pos);
        cnt -= !(--num[a[l++].id]);
    }
    cout << ans << endl;
    return 0;
}