[主席树][单调栈] BZOJ 4369: [IOI2015]teams分组

SolutionSolution

ki∈[Aj,Bj]ki∈[Aj,Bj]相当于点(Aj,Bj)(Aj,Bj)在(ki,ki)(ki,ki)的左上方。

那对于(ki,ki)(ki,ki)左上方的区域，有些点是在之前操作中已经被删掉的。

剩下的一些矩形的并。

画一下图，这些矩形的下边界是不升的，因为如果存在一对上升的下边界，那完全可以把大的那个删掉的点搞到前面去。

这样单调栈维护当前区域中点取完能不能够到kiki，以及如果剩下的点的数目在当前区域的下边界不满足单调性就把这个矩形合并掉。

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

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;
}