【CF833E】Caramel Clouds

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【CF833E】Caramel Clouds

题面

洛谷

题目大意：

天上有\(n\)朵云，每朵云\(i\)会在时间\([li,ri]\)出现，你有\(C\)个糖果，你可以花费\(c_i\)个糖果让云\(i\)消失，同时需要保证你最多让两朵云消失．现在有\(m\)个独立的询问，每次给你一个需要让阳光照\(k\)时间的植

物，问你从时刻\(0\)开始，这个植物最快什么时候能长成．

\(n,m\leq 3e5\)

题解

这题好神仙啊。。。

我们首先记几个东西：

\(Free:\)表示当前空着的长度和

\(single[x]:\)表示只有\(x\)覆盖的长度

\(opt[x]:\)选出的两朵云中必定包含\(x\)的最大长度

\(Top:\)\(opt\)的最大值

\(cross[x][y]:\)表示只有\(x,y\)覆盖的长度

然后用一个\(set\)维护当前还未消失的云

分类讨论一下\(set\)中云的个数：

1、当\(set\)中没有云时，直接让\(Free\)加上贡献即可

2、当\(set\)中只有一朵云，更新\(single\)以及\(opt\)

3、当\(set\)中有两朵云，更新\(cross\)、\(opt\)

4、当\(set\)中云朵数大于二，不用管，因为此时一定不会比云朵数小于等于二的时候答案更优

至于更新的方法，看代码应该能看懂

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <set>
#include <map>
using namespace std;
inline int gi() {
register int data = 0, w = 1;
register char ch = 0;
while (!isdigit(ch) && ch != '-') ch = getchar();
if (ch == '-') w = -1, ch = getchar();
while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();
return w * data;
}
void chkmax(int &x, int y) { if (x < y) x = y; }
const int MAX_N = 3e5 + 5;
struct Cloud { int l, r, c; } cld[MAX_N];
bool operator < (const Cloud &l, const Cloud &r) { return l.c < r.c; }
struct Query { int id, t; } q[MAX_N];
bool operator < (const Query &l, const Query &r) { return l.t < r.t; }
struct Node { int t, op, id; } a[MAX_N << 1]; int tot = 0;
bool operator < (const Node &l, const Node &r) { return l.t < r.t; }
int N, M, C, single[MAX_N], opt[MAX_N], ans[MAX_N];
map<int, int> cross[MAX_N];
set<int> s;
set<int>::iterator ite;
int Free, Top;
#define lson (o << 1)
#define rson (o << 1 | 1)
int mx[MAX_N << 2];
void modify(int o, int l, int r, int pos, int v) {
if (l == r) return (void)(mx[o] = v);
int mid = (l + r) >> 1;
if (pos <= mid) modify(lson, l, mid, pos, v);
else modify(rson, mid + 1, r, pos, v);
mx[o] = max(mx[lson], mx[rson]);
}
int find(int o, int l, int r) {
if (l == r) return l;
int mid = (l + r) >> 1;
if (mx[lson] > mx[rson]) return find(lson, l, mid);
else return find(rson, mid + 1, r);
}
int query(int o, int l, int r, int ql, int qr) {
if (ql <= l && r <= qr) return find(o, l, r);
int mid = (l + r) >> 1, res = 0;
if (ql <= mid) res = query(lson, l, mid, ql, qr);
if (qr > mid) {
int tmp = query(rson, mid + 1, r, ql, qr);
if (single[tmp] > single[res]) res = tmp;
}
return res;
}
int sum(int x, int y) {
if (x > y) swap(x, y);
return single[x] + single[y] + cross[x][y];
}
void solve() {
int now = 0, pos = 1;
for (int i = 1; i <= tot; i++) {
int dlt = a[i].t - now; now = a[i].t;
if (!s.size()) Free += dlt;
else if (s.size() == 1) {
ite = s.upper_bound(0); int x = *ite;
single[x] += dlt; modify(1, 1, N, x, single[x]);
opt[x] += dlt;
int rem = C - cld[x].c;
if (rem >= 0) {
int val = single[x];
if (rem >= cld[1].c) {
int l = 1, r = N, res = 1;
while (l <= r) {
int mid = (l + r) >> 1;
if (cld[mid].c <= rem) res = mid, l = mid + 1;
else r = mid - 1;
}
int ql = 1, qr = res;
if (x == qr) --qr;
if (x < qr) { ql = x + 1; if (ql <= qr) chkmax(val, sum(x, query(1, 1, N, ql, qr))); ql = 1, qr = x - 1; }
if (ql <= qr) chkmax(val, sum(x, query(1, 1, N, ql, qr)));
}
chkmax(opt[x], val);
chkmax(Top, opt[x]);
}
} else if (s.size() == 2) {
ite = s.upper_bound(0); int x = *ite;
++ite; int y = *ite;
if (cross[x].count(y) > 0) cross[x][y] += dlt;
else cross[x][y] = dlt;
if (cld[x].c + cld[y].c <= C) {
chkmax(opt[x], sum(x, y)); chkmax(opt[y], sum(x, y));
chkmax(Top, opt[x]);
}
}
while (pos <= M && Top + Free >= q[pos].t) ans[q[pos].id] = now - (Top + Free - q[pos].t), ++pos;
if (pos > M) break;
if (a[i].op == 1) s.insert(a[i].id);
else s.erase(a[i].id);
}
}
int main () {
N = gi(), C = gi();
for (int i = 1; i <= N; i++) cld[i].l = gi(), cld[i].r = gi(), cld[i].c = gi();
sort(&cld[1], &cld[N + 1]);
for (int i = 1; i <= N; i++) {
a[++tot] = (Node){cld[i].l, 1, i};
a[++tot] = (Node){cld[i].r, -1, i};
}
sort(&a[1], &a[tot + 1]);
a[++tot] = (Node){(int)(2e9 + 7), 1, N + 1};
M = gi();
for (int i = 1; i <= M; i++) q[i].id = i, q[i].t = gi();
sort(&q[1], &q[M + 1]);
solve();
for (int i = 1; i <= M; i++) printf("%d\n", ans[i]);
return 0;
}