9.5训练赛 D: Gym 102152B

思路：
差分对区间进行$O(n)$处理标记区间，分块取出每一块的数量和边界$r$，对每一块的数量进行排序，二分出符合x询问的边界值，建立线段树对该区间进行最大值询问。

注意：写线段树时，注意$n=0$

参考代码：

/*
 * @Author: vain
 * @Date: 2020-08-26 11:58:46
 * @LastEditTime: 2020-09-06 10:33:18
 * @LastEditors: sueRimn
 * @Description: In User Settings Edit
 * @FilePath: \main\demo.cpp
 */
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <ctime>
#include <cstring>
#include <cstdlib>
#include <math.h>
#include <bitset>
using namespace std;
typedef long long ll;
//#define ll long long
typedef unsigned long long ull;
const ll N = 2e3 + 20;
const ll maxn = 1e5 + 20;
const ll mod = 1e9 + 7;
ll head[maxn], stac[maxn];
int a[maxn], b[maxn], c[maxn];
//typedef pair<ll, ll> p;
//priority_queue<p, vector<p>, greater<p> > m;
//ll sum[maxn];
int max(int a, int b) { return a > b ? a : b; }
ll min(ll a, ll b) { return a < b ? a : b; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a * b / gcd(a, b); }
void swap(ll &x, ll &y) { x ^= y, y ^= x, x ^= y; }
map<string, ll> pll, che, mp;
ll f[N][3], q[N];
int lowbit(int x) { return (x) & (-x); }
vector<int> fc;
//ull h[maxn], p[maxn];
const ull base = 13331;
//char s1[maxn];
ll ksm(ll a, ll b)
{
    ll res = 1;
    while (b)
    {
        if (b & 1)
            res = a * res;
        b >>= 1;
        a = a * a;
    }
    return res;
}
struct node
{
    int r, sizx;
} p[maxn];
bool cmp(node x, node y)
{
    return x.sizx < y.sizx;
}
struct vain
{
    int l, r, ma;
} tr[maxn << 2];
int pushup(int k)
{
    tr[k].ma = max(tr[k << 1].ma, tr[k << 1 | 1].ma);
}
void build(int k, int l, int r)
{
    tr[k].l = l, tr[k].r = r;
    if (l == r)
    {
        tr[k].ma = p[l].r;
        return;
    }
    int mid = l + r >> 1;
    build(k << 1, l, mid);
    build(k << 1 | 1, mid + 1, r);
    pushup(k);
}
int query(int k, int ql, int qr)
{
    if (tr[k].l >= ql && tr[k].r <= qr)
    {
        return tr[k].ma;
    }
    int mid = tr[k].l + tr[k].r >> 1;
    int ans = 0;
    if (ql <= mid)
    {
        ans = max(ans, query(k << 1, ql, qr));
    }
    if (qr > mid)
    {
        ans = max(ans, query(k << 1 | 1, ql, qr));
    }
    return ans;
}
int main()
{
    // ios::sync_with_stdio(false);
    // cin.tie(0), cout.tie(0);
    // srand(time(NULL));
    int n, m, t, s, q;
    scanf("%d", &t);
    while (t--)
    {
        fc.clear();
        scanf("%d %d %d", &n, &m, &q);
        for (int i = 0; i <= m; i++)
            a[i] = 0;
        while (n--)
        {
            int l, r;
            scanf("%d %d", &l, &r);
            a[l]--;
            a[r + 1]++;
        }
        for (int i = 1; i <= m; i++)
            a[i] = a[i] + a[i - 1];
        int cnt = 0, ans = 0;
        for (int i = 1; i <= m; i++)
        {
            if (!a[i])
                ans++;
            else if (ans)
                p[++cnt].sizx = ans, p[cnt].r = i - 1, ans = 0;
        }
        if (ans)
            p[++cnt].sizx = ans, p[cnt].r = m;
        sort(p + 1, p + 1 + cnt, cmp);
        for (int i = 1; i <= cnt; i++)
            fc.push_back(p[i].sizx);
        if (cnt > 0)
            build(1, 1, cnt);
        while (q--)
        {
            int x;
            scanf("%d", &x);
            if (x > p[cnt].sizx)
                printf("-1 -1\n");
            else
            {
                int y = lower_bound(fc.begin(), fc.end(), x) - fc.begin() + 1;
                int qr = query(1, y, cnt);
                printf("%d %d\n", qr - x + 1, qr);
            }
        }
    }
}