CF1165F2(二分答案)

 题意:有i种物品，每种物品需要买k[i]个，然后商店会有m次特价出售，第j次为在di天出售第ti种物品，物品原价2元，特价1元，你每天上午可以获得一元，下午可以进行交易，求获取所有物品花费的最少时间为多少天？

思路:注意到,第i+1天可以,那么第i天一定可以,有单调性,所以使用二分答案.具体来说,二分日期x,在x内对每件商品,在其在x天内的最后时间用所有的钱买他,剩下的按照普通价买即可.

代码:

#include 
#define int long long
#define IOS ios::sync_with_stdio(false), cin.tie(0) 
#define ll long long 
// #define double long double
#define ull unsigned long long 
#define PII pair 
#define PDI pair 
#define PDD pair 
#define debug(a) cout << #a << " = " << a << endl 
#define point(n) cout << fixed << setprecision(n)
#define all(x) (x).begin(), (x).end() 
#define mem(x, y) memset((x), (y), sizeof(x)) 
#define lbt(x) (x & (-x)) 
#define SZ(x) ((x).size()) 
#define inf 0x3f3f3f3f 
#define INF 0x3f3f3f3f3f3f3f3f
namespace nqio{const unsigned R = 4e5, W = 4e5; char *a, *b, i[R], o[W], *c = o, *d = o + W, h[40], *p = h, y; bool s; struct q{void r(char &x){x = a == b && (b = (a = i) + fread(i, 1, R, stdin), a == b) ? -1 : *a++;} void f(){fwrite(o, 1, c - o, stdout); c = o;} ~q(){f();}void w(char x){*c = x;if (++c == d) f();} q &operator >>(char &x){do r(x);while (x <= 32); return *this;} q &operator >>(char *x){do r(*x); while (*x <= 32); while (*x > 32) r(*++x); *x = 0; return *this;} template<typename t> q&operator>>(t &x){for (r(y),s = 0; !isdigit(y); r(y)) s |= y == 45;if (s) for (x = 0; isdigit(y); r(y)) x = x * 10 - (y ^ 48); else for (x = 0; isdigit(y); r(y)) x = x * 10 + (y ^ 48); return *this;} q &operator <<(char x){w(x);return *this;}q &operator<< (char *x){while (*x) w(*x++); return *this;}q &operator <<(const char *x){while (*x) w(*x++); return *this;}template<typename t> q &operator<< (t x) {if (!x) w(48); else if (x < 0) for (w(45); x; x /= 10) *p++ = 48 | -(x % 10); else for (; x; x /= 10) *p++ = 48 | x % 10; while (p != h) w(*--p);return *this;}}qio; }using nqio::qio;
using namespace std;
const int N = 2e6 + 10;
int n, m, t[N];
vector<int> d[N], last[N], k;//last[i]表示在x天内最后销售时限为第i天的商品
bool check(int x) {
	auto s = k;
	for (int i = 1; i <= n; ++i) last[i].clear();
	for (int i = 1; i <= n; ++i) {
		int pos = 0;
		for (int xx : d[i]) {
			if (xx > x) break;
			pos = max(pos, xx);
		}
		if (pos) last[pos].emplace_back(i);
	}
	int mon = 0;
	for (int i = 1; i <= x; ++i) {
		++mon;
		for (int xx : last[i]) {
			if (mon > s[xx]) mon -= s[xx], s[xx] = 0;
			else s[xx] -= mon, mon = 0;
		}
	}
	for (int i = 1; i <= n; ++i) 
		mon -= (s[i] << 1);
	return mon >= 0;
}
void solve() {
	qio >> n >> m;
	k = vector<int> (n + 1, 0);
	for (int i = 1; i <= n; ++i) qio >> k[i];
	for (int i = 1, x; i <= m; ++i) {
		qio >> x >> t[i];
		d[t[i]].emplace_back(x);
	}
	for (int i = 1; i <= n; ++i) sort(all(d[i]));
	int l = 0, r = 1e6;
	while (l < r) {
		int mid = l + r >> 1;
		if (check(mid)) r = mid;
		else l = mid + 1;
	}
	qio << l << "\n";
	
}
signed main()  {
	int T = 1;
	// qio >> T;
	while (T--) solve();
}