[主席樹][單調棧] BZOJ 4369: [IOI2015]teams分組

Solution$Solution$

ki∈[Aj,Bj]$k_i\in [A_j,B_j]$ 相當於點(Aj,Bj)$(A_j,B_j)$ 在(ki,ki)$(k_i,k_i)$ 的左上方。
那對於(ki,ki)$(k_i,k_i)$ 左上方的區域，有些點是在之前操作中已經被刪掉的。
剩下的一些矩形的並。
畫一下圖，這些矩形的下邊界是不升的，因爲如果存在一對上升的下邊界，那完全可以把大的那個刪掉的點搞到前面去。
這樣單調棧維護當前區域中點取完能不能夠到ki$k_i$ ，以及如果剩下的點的數目在當前區域的下邊界不滿足單調性就把這個矩形合併掉。

#include <bits/stdc++.h>
#define show(x) cerr << #x << " = " << x << endl
using namespace std;
typedef long long ll;
typedef pair<int, int> Pairs;

const int N = 505050;

inline char get(void) {
    static char buf[100000], *S = buf, *T = buf;
    if (S == T) {
        T = (S = buf) + fread(buf, 1, 100000, stdin);
        if (S == T) return EOF;
    }
    return *S++;
}
template<typename T>
inline void read(T &x) {
    static char c; x = 0; int sgn = 0;
    for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 1;
    for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
    if (sgn) x = -x;
}

int a[N], b[N], k[N];
int h[N], sta[N], rest[N];
int n, q, m, tcnt, top;
int size[N * 70], ls[N * 70], rs[N * 70];
int rt[N];
vector<int> v[N];

inline void insert(int &o, int l, int r, int pos) {
    size[++tcnt] = size[o] + 1;
    ls[tcnt] = ls[o]; rs[tcnt] = rs[o];
    o = tcnt;
    if (l == r) return;
    int mid = (l + r) >> 1;
    if (pos <= mid) insert(ls[o], l, mid, pos);
    else insert(rs[o], mid + 1, r, pos);
}
inline int query(int x, int y, int pos) {
    int l = 1, r = n, res = 0;
    while (true) {
        if (l == r)
            return res += size[y] - size[x];
        int mid = (l + r) >> 1;
        if (pos <= mid) {
            res += size[rs[y]] - size[rs[x]];
            r = mid;
            x = ls[x]; y = ls[y];
        } else {
            l = mid + 1;
            x = rs[x]; y = rs[y];
        }
    }
}
inline int kth(int x, int y, int k) {
    int l = 1, r = n;
    while (true) {
        if (l == r) return l;
        int mid = (l + r) >> 1;
        if (k <= size[rs[y]] - size[rs[x]]) {
            l = mid + 1;
            x = rs[x]; y = rs[y];
        } else {
            k -= size[rs[y]] - size[rs[x]];
            r = mid;
            x = ls[x]; y = ls[y];
        }
    }
}

int main(void) {
    read(n);
    for (int i = 1; i <= n; i++) {
        read(a[i]); read(b[i]);
        v[a[i]].push_back(b[i]);
    }
    for (int i = 1; i <= n; i++) {
        rt[i] = rt[i - 1];
        for (int j = 0; j < v[i].size(); j++)
            insert(rt[i], 1, n, v[i][j]);
    }
    read(q);
    while (q--) {
        read(m); top = 0;
        for (int i = 1; i <= m; i++) read(k[i]);
        sort(k + 1, k + m + 1);
        for (int i = 1; i <= m; i++) {
            if (k[i] > n) {
                printf("0\n"); break;
            }
            while (top && h[top] < k[i]) --top;
            int _rest = rest[top], _h;
            _rest += query(rt[k[sta[top]]], rt[k[i]], k[i]) - k[i];
            if (_rest < 0) {
                printf("0\n"); break;
            } else if (i == m) {
                printf("1\n"); break;
            }
            while ((_h = kth(rt[k[sta[top]]], rt[k[i]], _rest - rest[top])) > h[top] && top) --top;
            sta[++top] = i; rest[top] = _rest; h[top] = _h;
        }
    }
    return 0;
}