JZOJ 100149. 一道联赛A题

\(\text{Solution}\)

一眼 \(ODT\)
为避免每次都数颜色数量,提前记录下来,每次修改更新下

\(\text{Code}\)

#include <cstdio> 
#include <iostream>
#include <set>
#define re register
using namespace std;

const int N = 1e5 + 5;
int L, c, m, cnt[N];

inline void read(int &x)
{
	x = 0; char ch = getchar();
	for(; !isdigit(ch); ch = getchar());
	for(; isdigit(ch); x = (x<<3) + (x<<1) + (ch^48), ch = getchar());
}

struct node{
	int l, r; mutable int v;
	inline node(int l, int r, int v):l(l), r(r), v(v){};
	inline bool operator < (const node &a) const {return l < a.l;}
};
typedef set<node>::iterator iter;
set<node> T;
inline iter split(int x)
{
	if (x >= L) return T.end();
	iter it = --T.upper_bound(node{x, 0, 0});
	if (it->l == x) return it;
	int l = it->l, r = it->r, v = it->v;
	T.erase(it), T.insert(node{l, x - 1, v});
	return T.insert(node{x, r, v}).first;
}
inline void assign(int l, int r, int v)
{
	iter itr = split(r + 1), itl = split(l);
	for(re iter it = itl; it != itr; ++it) cnt[it->v] -= (it->r - it-> l + 1);
	T.erase(itl, itr), cnt[v] += r - l + 1, T.insert(node{l, r, v});
}

int main() 
{
	read(L), read(c), read(m), T.insert(node{0, L - 1, 1}), cnt[1] = L;
	for(int p, x, a, b, s, l, r; m; --m)
	{
		read(p), read(x), read(a), read(b), s = cnt[p];
		l = (a + (long long)s * s) % L, r = (a + (long long)(s + b) * (s + b)) % L;
		if (l > r) swap(l, r); assign(l, r, x);
	}
	int mx = 0;
	for(re int i = 1; i <= c; i++) mx = (cnt[i] > mx ? cnt[i] : mx);
	printf("%d\n", mx);
}
posted @ 2021-11-14 20:08  leiyuanze  阅读(18)  评论(0编辑  收藏  举报