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题解:P10639 BZOJ4695 最佳女选手

时间:2024-06-20 23:32:14浏览次数:16  
标签:return 题解 ll BZOJ4695 tag m1 女选手 Modint ql

区间最值操作基础题,但是有点码农。

依然考虑势能线段树,维护区间和 \(\textrm{sum}\)、最大值 \(\textrm{M1}\)、次大值 \(\textrm{M2}\)、最大值个数 \(\textrm{Mcnt}\)、最小值 \(\textrm{m1}\)、次小值 \(\textrm{m2}\)、最小值个数 \(\textrm{mcnt}\),另外需要区间加标记 \(\textrm{tags}\)、区间取最大值标记 \(\textrm{tagM}\)、区间取最小值标记 \(\textrm{tagm}\)。约定当区间只包含一个元素时,有 \(\textrm{M2}=-\infty,\textrm{m2}=+\infty\),且初始情况下 \(\textrm{tagM}=-\infty,\textrm{tagm}=+\infty\)。

鉴于需要维护这么多信息和操作,本题的一个难点在于下传标记时的顺序需要正确实现。我的势能线段树的标记优先级为:区间加标记最优先,两种区间最值标记优先级相同。

在下传标记时,我们先下传区间加标记,同时对应地修改区间最值标记,再使用修改后的区间最值标记进行下传即可。由于势能线段树只会在仅有最大值/最小值发生变化时下传标记,因此这一部分容易实现。

可以证明时间复杂度为 \(O(n\log^2n)\)。

// Problem: P10639 BZOJ4695 最佳女选手
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P10639
// Memory Limit: 512 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

//By: OIer rui_er
#include <bits/stdc++.h>
#define rep(x, y, z) for(ll x = (y); x <= (z); ++x)
#define per(x, y, z) for(ll x = (y); x >= (z); --x)
#define debug(format...) fprintf(stderr, format)
#define fileIO(s) do {freopen(s".in", "r", stdin); freopen(s".out", "w", stdout);} while(false)
#define endl '\n'
using namespace std;
typedef long long ll;

mt19937 rnd(std::chrono::duration_cast<std::chrono::nanoseconds>(std::chrono::system_clock::now().time_since_epoch()).count());
int randint(int L, int R) {
    uniform_int_distribution<int> dist(L, R);
    return dist(rnd);
}

template<typename T> void chkmin(T& x, T y) {if(x > y) x = y;}
template<typename T> void chkmax(T& x, T y) {if(x < y) x = y;}

template<int mod>
inline unsigned int down(unsigned int x) {
	return x >= mod ? x - mod : x;
}

template<int mod>
struct Modint {
	unsigned int x;
	Modint() = default;
	Modint(unsigned int x) : x(x) {}
	friend istream& operator>>(istream& in, Modint& a) {return in >> a.x;}
	friend ostream& operator<<(ostream& out, Modint a) {return out << a.x;}
	friend Modint operator+(Modint a, Modint b) {return down<mod>(a.x + b.x);}
	friend Modint operator-(Modint a, Modint b) {return down<mod>(a.x - b.x + mod);}
	friend Modint operator*(Modint a, Modint b) {return 1ULL * a.x * b.x % mod;}
	friend Modint operator/(Modint a, Modint b) {return a * ~b;}
	friend Modint operator^(Modint a, int b) {Modint ans = 1; for(; b; b >>= 1, a *= a) if(b & 1) ans *= a; return ans;}
	friend Modint operator~(Modint a) {return a ^ (mod - 2);}
	friend Modint operator-(Modint a) {return down<mod>(mod - a.x);}
	friend Modint& operator+=(Modint& a, Modint b) {return a = a + b;}
	friend Modint& operator-=(Modint& a, Modint b) {return a = a - b;}
	friend Modint& operator*=(Modint& a, Modint b) {return a = a * b;}
	friend Modint& operator/=(Modint& a, Modint b) {return a = a / b;}
	friend Modint& operator^=(Modint& a, int b) {return a = a ^ b;}
	friend Modint& operator++(Modint& a) {return a += 1;}
	friend Modint operator++(Modint& a, int) {Modint x = a; a += 1; return x;}
	friend Modint& operator--(Modint& a) {return a -= 1;}
	friend Modint operator--(Modint& a, int) {Modint x = a; a -= 1; return x;}
	friend bool operator==(Modint a, Modint b) {return a.x == b.x;}
	friend bool operator!=(Modint a, Modint b) {return !(a == b);}
};

const ll N = 5e5 + 5, inf = 0x3f3f3f3f3f3f3f3fll;

ll n, m, a[N];

struct SegTree {
    ll sum[N << 2], M1[N << 2], M2[N << 2], Mcnt[N << 2], m1[N << 2], m2[N << 2], mcnt[N << 2];
    ll tagM[N << 2], tagm[N << 2], tags[N << 2];
    #define lc(u) (u << 1)
    #define rc(u) (u << 1 | 1)
    void pushup(ll u) {
        sum[u] = sum[lc(u)] + sum[rc(u)];
        M1[u] = max(M1[lc(u)], M1[rc(u)]);
        M2[u] = -inf;
        if(M1[lc(u)] < M1[u]) chkmax(M2[u], M1[lc(u)]);
        if(M2[lc(u)] < M1[u]) chkmax(M2[u], M2[lc(u)]);
        if(M1[rc(u)] < M1[u]) chkmax(M2[u], M1[rc(u)]);
        if(M2[rc(u)] < M1[u]) chkmax(M2[u], M2[rc(u)]);
        Mcnt[u] = 0;;
        if(M1[lc(u)] == M1[u]) Mcnt[u] += Mcnt[lc(u)];
        if(M1[rc(u)] == M1[u]) Mcnt[u] += Mcnt[rc(u)];
        m1[u] = min(m1[lc(u)], m1[rc(u)]);
        m2[u] = +inf;
        if(m1[lc(u)] > m1[u]) chkmin(m2[u], m1[lc(u)]);
        if(m2[lc(u)] > m1[u]) chkmin(m2[u], m2[lc(u)]);
        if(m1[rc(u)] > m1[u]) chkmin(m2[u], m1[rc(u)]);
        if(m2[rc(u)] > m1[u]) chkmin(m2[u], m2[rc(u)]);
        mcnt[u] = 0;;
        if(m1[lc(u)] == m1[u]) mcnt[u] += mcnt[lc(u)];
        if(m1[rc(u)] == m1[u]) mcnt[u] += mcnt[rc(u)];
    }
    void pushs(ll u, ll l, ll r, ll tag) {
        sum[u] += tag * (r - l + 1);
        if(M1[u] != -inf) M1[u] += tag;
        if(M2[u] != -inf) M2[u] += tag;
        if(m1[u] != +inf) m1[u] += tag;
        if(m2[u] != +inf) m2[u] += tag;
        tags[u] += tag;
        if(tagM[u] != -inf) tagM[u] += tag;
        if(tagm[u] != +inf) tagm[u] += tag;
    }
    void pushM(ll u, ll l, ll r, ll tag) {
        if(m1[u] > tag) return;
        sum[u] += (tag - m1[u]) * mcnt[u];
        if(M2[u] == m1[u]) M2[u] = tag;
        if(M1[u] == m1[u]) M1[u] = tag;
        m1[u] = tag;
        chkmax(tagM[u], tag);
        chkmax(tagm[u], tag);
    }
    void pushm(ll u, ll l, ll r, ll tag) {
        if(M1[u] < tag) return;
        sum[u] += (tag - M1[u]) * Mcnt[u];
        if(m2[u] == M1[u]) m2[u] = tag;
        if(m1[u] == M1[u]) m1[u] = tag;
        M1[u] = tag;
        chkmin(tagM[u], tag);
        chkmin(tagm[u], tag);
    }
    void pushdown(ll u, ll l, ll r) {
        ll mid = (l + r) >> 1;
        if(tags[u]) {
            pushs(lc(u), l, mid, tags[u]);
            pushs(rc(u), mid + 1, r, tags[u]);
            tags[u] = 0;
        }
        if(tagM[u] != -inf) {
            pushM(lc(u), l, mid, tagM[u]);
            pushM(rc(u), mid + 1, r, tagM[u]);
            tagM[u] = -inf;
        }
        if(tagm[u] != +inf) {
            pushm(lc(u), l, mid, tagm[u]);
            pushm(rc(u), mid + 1, r, tagm[u]);
            tagm[u] = +inf;
        }
    }
    void build(ll u, ll l, ll r) {
        tags[u] = 0;
        tagM[u] = -inf;
        tagm[u] = +inf;
        if(l == r) {
            sum[u] = M1[u] = m1[u] = a[l];
            M2[u] = -inf;
            m2[u] = +inf;
            Mcnt[u] = mcnt[u] = 1;
            return;
        }
        ll mid = (l + r) >> 1;
        build(lc(u), l, mid);
        build(rc(u), mid + 1, r);
        pushup(u);
    }
    void modifys(ll u, ll l, ll r, ll ql, ll qr, ll k) {
        if(ql <= l && r <= qr) {
            pushs(u, l, r, k);
            return;
        }
        pushdown(u, l, r);
        ll mid = (l + r) >> 1;
        if(ql <= mid) modifys(lc(u), l, mid, ql, qr, k);
        if(qr > mid) modifys(rc(u), mid + 1, r, ql, qr, k);
        pushup(u);
    }
    void modifyM(ll u, ll l, ll r, ll ql, ll qr, ll k) {
        if(m1[u] >= k) return;
        if(ql <= l && r <= qr && m2[u] > k) {
            pushM(u, l, r, k);
            return;
        }
        pushdown(u, l, r);
        ll mid = (l + r) >> 1;
        if(ql <= mid) modifyM(lc(u), l, mid, ql, qr, k);
        if(qr > mid) modifyM(rc(u), mid + 1, r, ql, qr, k);
        pushup(u);
    }
    void modifym(ll u, ll l, ll r, ll ql, ll qr, ll k) {
        if(M1[u] <= k) return;
        if(ql <= l && r <= qr && M2[u] < k) {
            pushm(u, l, r, k);
            return;
        }
        pushdown(u, l, r);
        ll mid = (l + r) >> 1;
        if(ql <= mid) modifym(lc(u), l, mid, ql, qr, k);
        if(qr > mid) modifym(rc(u), mid + 1, r, ql, qr, k);
        pushup(u);
    }
    ll querys(ll u, ll l, ll r, ll ql, ll qr) {
        if(ql <= l && r <= qr) return sum[u];
        pushdown(u, l, r);
        ll mid = (l + r) >> 1, ans = 0;
        if(ql <= mid) ans += querys(lc(u), l, mid, ql, qr);
        if(qr > mid) ans += querys(rc(u), mid + 1, r, ql, qr);
        pushup(u);
        return ans;
    }
    ll queryM(ll u, ll l, ll r, ll ql, ll qr) {
        if(ql <= l && r <= qr) return M1[u];
        pushdown(u, l, r);
        ll mid = (l + r) >> 1, ans = -inf;
        if(ql <= mid) chkmax(ans, queryM(lc(u), l, mid, ql, qr));
        if(qr > mid) chkmax(ans, queryM(rc(u), mid + 1, r, ql, qr));
        pushup(u);
        return ans;
    }
    ll querym(ll u, ll l, ll r, ll ql, ll qr) {
        if(ql <= l && r <= qr) return m1[u];
        pushdown(u, l, r);
        ll mid = (l + r) >> 1, ans = +inf;
        if(ql <= mid) chkmin(ans, querym(lc(u), l, mid, ql, qr));
        if(qr > mid) chkmin(ans, querym(rc(u), mid + 1, r, ql, qr));
        pushup(u);
        return ans;
    }
    #undef lc
    #undef rc
}sgt;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    cin >> n;
    rep(i, 1, n) cin >> a[i];
    sgt.build(1, 1, n);
    for(cin >> m; m; --m) {
        ll op;
        cin >> op;
        if(op == 1) {
            ll l, r, x;
            cin >> l >> r >> x;
            sgt.modifys(1, 1, n, l, r, x);
        }
        else if(op == 2) {
            ll l, r, x;
            cin >> l >> r >> x;
            sgt.modifyM(1, 1, n, l, r, x);
        }
        else if(op == 3) {
            ll l, r, x;
            cin >> l >> r >> x;
            sgt.modifym(1, 1, n, l, r, x);
        }
        else if(op == 4) {
            ll l, r;
            cin >> l >> r;
            cout << sgt.querys(1, 1, n, l, r) << endl;
        }
        else if(op == 5) {
            ll l, r;
            cin >> l >> r;
            cout << sgt.queryM(1, 1, n, l, r) << endl;
        }
        else {
            ll l, r;
            cin >> l >> r;
            cout << sgt.querym(1, 1, n, l, r) << endl;
        }
    }
    return 0;
}

标签:return,题解,ll,BZOJ4695,tag,m1,女选手,Modint,ql
From: https://www.cnblogs.com/ruierqwq/p/18259675/LG-P10639

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