• 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);
    }
    
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  • 原文地址:https://www.cnblogs.com/leiyuanze/p/15553078.html
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