CDQ分治

什么是CDQ分治？

一个最简单的cdq分治的例子就是归并排序求逆序对。

简单的说，cdq分治就是有一系列关于区间$[L,R]$的问题。

1.递归处理$[L,M]$和$[M+1,R]$。

2.计算$[L,M]$对$[M+1,R]$的影响。

1176: [Balkan2007]Mokia

1790: [Ahoi2008]Rectangle 矩形藏宝地

按时间$T$排序，在$solve$中再按$x$排序，然后用数据结构维护$y$即可。

#include<cstdio>
#include<iostream>
using namespace std;
inline char nc() {
    static char b[1<<16],*s=b,*t=b;
    return s==t&&(t=(s=b)+fread(b,1,1<<16,stdin),s==t)?-1:*s++;
}
inline void read(int &x) {
    char b = nc(); x = 0;
    for (; !isdigit(b); b = nc());
    for (; isdigit(b); b = nc()) x = x * 10 + b - '0';
}
int s, n, ans[10010], m, c[2000005];
inline int lowbit(int x) {
    return x & -x;  
}
inline void add(int p, int x) {
    for (; p <= n; p += lowbit(p)) c[p] += x;    
}
inline int sum(int p) {
    int res = 0;
    for (; p; p -= lowbit(p)) res += c[p];
    return res;
}
struct Q {
    int x, y, v, T, a;
    inline bool operator<(const Q &q) const {
        return x != q.x ? x < q.x : T < q.T;
    }
} q[200005], tq[200005];
inline void add(int x, int y, int v, int a) {
    q[++m] = (Q){x, y, v, m, a};
}
void solve(int l, int r) {
    if (l == r) return;
    int m = (l + r) >> 1, h1 = l, h2 = m + 1, h = l;
    solve(l, m); solve(m + 1, r);
    while (h1 <= m && h2 <= r) tq[h++] = q[h1] < q[h2] ? q[h1++] : q[h2++];
    while (h1 <= m) tq[h++] = q[h1++]; while (h2 <= r) tq[h++] = q[h2++];
    for (int i = l; i <= r; ++i) q[i] = tq[i];
    for (int i = l; i <= r; ++i) {
        if (!q[i].a && q[i].T <= m) add(q[i].y, q[i].v);
        if (q[i].a && q[i].T > m) ans[q[i].a] += q[i].v * sum(q[i].y);
    }
    for (int i = l; i <= r; ++i)
        if (!q[i].a && q[i].T <= m) add(q[i].y, -q[i].v);
}
int main() {
    read(s); read(n); ++n;
    for (int op, a, b, c, d;;) {
        read(op); if (op == 3) break;
        read(a), read(b), read(c);
        if (op == 1) add(a + 1, b + 1, c, 0);
        else if (op == 2) {
            read(d); ++a; ++b; ++c; ++d; ++ans[0];
            add(c, d, 1, ans[0]);
            add(a - 1, b - 1, 1, ans[0]);
            add(a - 1, d, -1, ans[0]);
            add(c, b - 1, -1, ans[0]);
        }
    }
    solve(1, m);
    for (int i = 1; i <= ans[0]; ++i) printf("%d\n", ans[i]);
    return 0;
}

1492: [NOI2007]货币兑换Cash

每次买入卖出都用完钱或者券是最吼的。

记$dp_i$，$f_i$，$g_i$为第$i$天最多的钱，A券，B券。

由题意可知，$f_i=dp_i\frac{R_i}{A_iR_i+B_i}$，$g_i=\frac{R_i}{A_iR_i+B_i}$。

$dp$方程很显然，$dp_i=\max\{A_if_j + B_ig_j\}(j < i)$

这样我们就得到了一个$O(N^2)$的算法，但是还不行。

若对于$i$，决策$j$比$k$更优，则，

$A_if_j+B_ig_j>A_if_k+B_ig_k$

$A_i(f_j-f_k)+B_i(g_j-g_k)>0$

为了处理符号，咱不妨设$f_j<f_k$，

$\frac{g_j-g_k}{f_j-f_k}<-\frac{A_i}{B_i}$

这就是一个斜率优化的式子了，但是左右两边都没有单调性。

这时候就是cdq分治出场的时候了。

咱先将每一天按照$-\frac{A_i}{B_i}$排序，然后再在递归处理时按$i$排序，并维护一个斜率单调递减的下凸壳。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 100010;
typedef double db;
inline char nc() {
    static char b[1<<16],*s=b,*t=b;
    return s==t&&(t=(s=b)+fread(b,1,1<<16,stdin),s==t)?-1:*s++;
}
inline void read(int &x) {
    char b = nc(); x = 0;
    for (; !isdigit(b); b = nc());
    for (; isdigit(b); b = nc()) x = x * 10 + b - '0';
}
inline void read(db &x) {
    char b = nc(); x = 0;
    for (; !isdigit(b); b = nc());
    for (; isdigit(b); b = nc()) x = x * 10 + b - '0';
    if (b == '.') {
        db t = 0.1;
        for (b = nc(); isdigit(b); b = nc()) 
            x += t * (b - '0'), t /= 10;
    }
}
int n, top, stk[N];
db dp[N];
struct Q {
    db a, b, r, s;
    int id;
    inline void init(int i) {
        read(a); read(b); read(r);
        s = -a / b; id = i;
    }
    inline bool operator<(const Q &q) const {
        return s < q.s;
    }
} q[N], tq[N];
struct Pt {
    db x, y;
    inline bool operator<(const Pt &p) const {
        return x != p.x ? x < p.x : y < p.y;    
    }
} p[N], tp[N];
bool check(const Pt &a, const Pt &b, const Pt &c) {
    return (b.y - a.y) * (c.x - b.x) > (b.x - a.x) * (c.y - b.y);
}
bool check(int a, int b, int c) {
    return check(p[a], p[b], p[c]); 
}
inline void gmax(db &x, db y) {
    if (x < y) x = y;    
}
db slope(const Pt &a, const Pt &b) {
    if (a.x == b.x) return a.y < b.y ? 1e10 : -1e10;
    return (a.y - b.y) / (a.x - b.x);
}
db slope(int a, int b) {
    return slope(p[a], p[b]);
}
void solve(int l, int r) {
    if (l == r) {
        gmax(dp[l], dp[l-1]);
        p[l].x = dp[l] * q[l].r / (q[l].a * q[l].r + q[l].b);
        p[l].y = dp[l] / (q[l].a * q[l].r + q[l].b);
        return;
    }
    int m = (l + r) >> 1, h1 = l, h2 = m + 1;
    for (int i = l; i <= r; ++i)
        if (q[i].id <= m) tq[h1++] = q[i];
        else tq[h2++] = q[i];
    for (int i = l; i <= r; ++i) q[i] = tq[i];
    solve(l, m); top = 0;
    for (int i = l; i <= m; ++i) {
        while (top > 1 && !check(stk[top-1], stk[top], i)) --top;
        stk[++top] = i;
    }
    for (int i = m + 1; i <= r; ++i) {
        while (top > 1 && slope(stk[top-1], stk[top]) < q[i].s) --top;
        gmax(dp[q[i].id], q[i].a * p[stk[top]].x + q[i].b * p[stk[top]].y);
    }
    solve(m + 1, r);
    h1 = l, h2 = m + 1;
    for (int i = l; i <= r; ++i)
        if (h1 <= m && (p[h1] < p[h2] || h2 > r)) tp[i] = p[h1++];
        else tp[i] = p[h2++];
    for (int i = l; i <= r; ++i) p[i] = tp[i];
}
inline void foo() {
    freopen("cash.in", "r", stdin);
    freopen("cash.out", "w", stdout);
}
int main() {
    read(n); read(dp[0]);
    for (int i = 1; i <= n; ++i) q[i].init(i);
    sort(q + 1, q + 1 + n); 
    solve(1, n); printf("%.3lf\n", dp[n]);
    return 0;
}

3262: 陌上花开

一道三维偏序问题。

都是套路。先在外面按$s$排序，递归时按$c$排序，树状数组维护$m$。

由于这题是$\leqslant$，所以还有一些额外的操作！

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
inline char nc() {
    static char b[1<<16],*s=b,*t=b;
    return s==t&&(t=(s=b)+fread(b,1,1<<16,stdin),s==t)?-1:*s++;
}
inline void read(int &x) {
    char b = nc(); x = 0;
    for (; !isdigit(b); b = nc());
    for (; isdigit(b); b = nc()) x = x * 10 + b - '0';
}
struct Node {
    int a, b, c, T, ans;
    inline void init() {
        read(a), read(b), read(c);
    }
    inline bool operator<(const Node &o) const {
        if (a != o.a) return a < o.a;
        if (b != o.b) return b < o.b;
        return c < o.c;
    }
    inline bool operator==(const Node &o) const {
        return a == o.a && b == o.b && c == o.c;
    }
} p[100005], tp[100005];
int n, H, c[200005], ans[100005];
inline int lowbit(int x) {
    return x & -x;
}
inline void add(int p, int x) {
    for (; p <= H; p += lowbit(p)) c[p] += x;    
}
inline int sum(int p) {
    int res = 0;
    for (; p; p -= lowbit(p)) res += c[p];
    return res;
}
void solve(int l, int r) {
    if (l == r) return;
    int m = (l + r) >> 1, h1 = l, h2 = m + 1, h = l;
    solve(l, m); solve(m + 1, r);
    for (int i = l; i <= m; ++i) p[i].T = 1;
    for (int i = m + 1; i <= r; ++i) p[i].T = 0;
    while (h1 <= m && h2 <= r) tp[h++] = p[h1].b <= p[h2].b ? p[h1++] : p[h2++];
    while (h1 <= m) tp[h++] = p[h1++]; while (h2 <= r) tp[h++] = p[h2++];
    for (int i = l; i <= r; ++i) p[i] = tp[i];
    for (int i = l; i <= r; ++i) {
        if (p[i].T) add(p[i].c, 1);
        else p[i].ans += sum(p[i].c);
    }
    for (int i = l; i <= r; ++i) if (p[i].T) add(p[i].c, -1);
}
 
int main() {
    read(n); read(H);
    for (int i = 1; i <= n; ++i) p[i].init();
    sort(p + 1, p + 1 + n); solve(1, n);
    for (int i = n - 1; i; --i) if (p[i] == p[i+1]) 
        p[i].ans = max(p[i].ans, p[i+1].ans);
    for (int i = 1; i <= n; ++i) ++ans[p[i].ans];
    for (int i = 0; i < n; ++i) printf("%d\n", ans[i]);
    return 0;   
}

2479. [HZOI 2016]偏序

一道四维偏序题。记四元组$(a,b,c,d)$。跟三维偏序的套路一样。我们先按$a$排序，第一层$cdq$分治按$b$排序，得到一些四元组$(L,b,c,d)$和$(R,b,c,d)$。虽然这样排序之后$a$是无序的，但是$(L,b,c,d)$的$a$与$(R,b,c,d)$的$a$的大小关系是可以确定的。第二层$cdq$分治中，咱再按$c$排序，得到一些四元组$(L/R,L/R,c,d)$，再用树状数组维护$d$。

回忆三维偏序，咱只更新$(L,b,c)$的$c$的个数，只更新$(R,b,c)$的答案。在四维偏序里也是类似的。咱只更新$(L,L,c,d)$的$d$的个数，只更新$(R,R,c,d)$的答案。

如果还没看明白，这里安利一发__stdcall的讲解。

#include<cstdio>
#include<iostream>
#define setO(p, l, r, o, v) for (int i = l; i <= r; ++i) p[i].o = v
using namespace std;
inline char nc() {
    static char b[1<<16],*s=b,*t=b;
    return s==t&&(t=(s=b)+fread(b,1,1<<16,stdin),s==t)?-1:*s++;
}
inline void read(int &x) {
    char b = nc(); x = 0;
    for (; !isdigit(b); b = nc());
    for (; isdigit(b); b = nc()) x = x * 10 + b - '0';
}
const int N = 50005;
int n, c[N];
unsigned int ans;
inline int lowbit(int x) {return x & -x;}
inline void add(int p, int x) {for (; p <= n; p += lowbit(p)) c[p] += x;}
inline int sum(int p) {
    int res = 0;
    for (; p; p -= lowbit(p)) res += c[p];
    return res;
}
struct P {
    int b, c, d; bool o1, o2;
    inline bool L() {return o1 && o2;}
    inline bool R() {return !o1 && !o2;}
} p[N], p1[N], p2[N];
void solve2(int l, int r) {
    if (l == r) return;
    int m = (l + r) >> 1, h1 = l, h2 = m + 1, h = l;
    solve2(l, m); solve2(m + 1, r);
    setO(p1, l, m, o2, 1); setO(p1, m + 1, r, o2, 0);
    while (h1 <= m && h2 <= r) p2[h++] = p1[h1].c < p1[h2].c ? p1[h1++] : p1[h2++];
    while (h1 <= m) p2[h++] = p1[h1++]; while (h2 <= r) p2[h++] = p1[h2++];
    for (int i = l; i <= r; ++i) p1[i] = p2[i];
    for (int i = l; i <= r; ++i) {
        if (p1[i].L()) add(p1[i].d, 1);
        else if (p1[i].R()) ans += sum(p1[i].d);
    }
    for (int i = l; i <= r; ++i) if (p1[i].L()) add(p1[i].d, -1);
}
void solve1(int l, int r) {
    if (l == r) return;
    int m = (l + r) >> 1, h1 = l, h2 = m + 1, h = l;
    solve1(l, m); solve1(m + 1, r);
    setO(p, l, m, o1, 1); setO(p, m + 1, r, o1, 0);
    while (h1 <= m && h2 <= r) p1[h++] = p[h1].b < p[h2].b ? p[h1++] : p[h2++];
    while (h1 <= m) p1[h++] = p[h1++]; while (h2 <= r) p1[h++] = p[h2++];
    for (int i = l; i <= r; ++i) p[i] = p1[i]; solve2(l, r);
}
int main() {
    freopen("partial_order.in", "r", stdin);
    freopen("partial_order.out", "w", stdout);
    read(n);
    for (int i = 1; i <= n; ++i) read(p[i].b);
    for (int i = 1; i <= n; ++i) read(p[i].c);
    for (int i = 1; i <= n; ++i) read(p[i].d);
    solve1(1, n); printf("%u\n", ans);
    return 0;    
}

2716: [Violet 3]天使玩偶

绝对值很麻烦，所以我们每次只更新在左下方的点。旋转坐标系，一共做四次就可以得到答案了。cdq分治并维护前缀最大值即可。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
inline char nc() {
    static char b[1<<16],*s=b,*t=b;
    return s==t&&(t=(s=b)+fread(b,1,1<<16,stdin),s==t)?-1:*s++;
}
inline void read(int &x) {
    char b = nc(); x = 0;
    for (; !isdigit(b); b = nc());
    for (; isdigit(b); b = nc()) x = x * 10 + b - '0';
}
struct P {
    int x, y, o, T;
    inline bool operator<(const P &p) const {
        return x != p.x ? x < p.x : T < p.T;  
    }
} p[1000005], tp[1000005], op[1000005];
int n, m, mx[1000010], xmax, ymax, A, ans[500005];
const int inf = 0x3f3f3f3f;
inline void gmax(int &x, int y) {if (x < y) x = y;}
inline void gmin(int &x, int y) {if (x > y) x = y;}
inline int lowbit(int x) {return x & -x;}
inline void add(int p, int x) {
    for (; p <= ymax; p += lowbit(p)) gmax(mx[p], x);    
}
inline int getmx(int p) {
    int res = 0;
    for (; p; p -= lowbit(p)) gmax(res, mx[p]);
    return res == 0 ? -inf : res;
}
inline void clrmx(int p) {
    for (; p <= ymax; p += lowbit(p)) mx[p] = 0; 
}
void solve(int l, int r) {
    if (l == r) return ;
    int m = (l + r) >> 1, h1 = l, h2 = m + 1, h = l;
    solve(l, m); solve(m + 1, r);
    while (h1 <= m && h2 <= r) tp[h++] = p[h1] < p[h2] ? p[h1++] : p[h2++];
    while (h1 <= m) tp[h++] = p[h1++]; while (h2 <= r) tp[h++] = p[h2++];
    for (int i = l; i <= r; ++i) p[i] = tp[i];
    for (int i = l; i <= r; ++i) {
        if (p[i].T <= m && !p[i].o) add(p[i].y, p[i].x + p[i].y);
        else if (p[i].T > m && p[i].o) gmin(ans[p[i].o], p[i].x + p[i].y - getmx(p[i].y));
    }
    for (int i = l; i <= r; ++i)
        if (p[i].T <= m && !p[i].o) clrmx(p[i].y);
}
void cpy() {
    for (int i = 1; i <= n + m; ++i) p[i] = op[i];
}
int main() {
    read(n); read(m);
    for (int x, y, i = 1; i <= n; ++i)
        read(x), read(y), op[i] = (P){x + 1, y + 1, 0, i};
    for (int x, y, o, i = n + 1; i <= n + m; ++i)
        read(o), read(x), read(y), op[i] = (P){x + 1, y + 1, o == 1 ? 0 : ++A, i};
    for (int i = 1; i <= A; ++i) ans[i] = inf;
    for (int i = 1; i <= n + m; ++i) gmax(xmax, op[i].x), gmax(ymax, op[i].y);
    ++xmax, ++ymax; cpy(); solve(1, n + m);
    cpy(); for (int i = 1; i <= n + m; ++i) p[i].x = -p[i].x + xmax; solve(1, n + m);
    cpy(); for (int i = 1; i <= n + m; ++i) p[i].y = -p[i].y + ymax; solve(1, n + m);
    cpy(); for (int i = 1; i <= n + m; ++i) p[i].x = -p[i].x + xmax, p[i].y = -p[i].y + ymax; solve(1, n + m);
    for (int i = 1; i <= A; ++i) printf("%d\n", ans[i]);
    return 0;
}

4170: 极光

把数列看成一个个点$(x, a[x])$，再把坐标系转$\pi/4$，问题转化成询问矩形内点的个数。

TBD: 

http://www.lydsy.com/JudgeOnline/problem.php?id=2001