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Section 4.4

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大蒟蒻 9 years ago
parent bb5db70ffd
commit 0050bc4d6a
8 changed files with 331 additions and 53 deletions
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53
OnlineJudges/poj/3468.cpp
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#include <cstdio>
typedef long long int64;
#define lch(x) ((x) << 1)
#define rch(x) ((x) << 1 | 1)
#define avg(x, y) (((x) + (y)) >> 1)
inline int min(int a, int b)
{
return a < b ? a : b;
}
inline int max(int a, int b) { return a > b ? a : b; }
const int N = 100010;
int64 sum[N << 2], add[N << 2], a[N];
void build(int k, int l, int r)
{
if (l + 1 == r) return void(sum[k] = a[l]);
int m = avg(l, r);
build(lch(k), l, m);
build(rch(k), m, r);
sum[k] += sum[lch(k)] + sum[rch(k)];
}
void modify(int k, int l, int r, int x, int y, int v)
{
if (x <= l && r <= y) return void(add[k] += v);
sum[k] += (min(r, y) - max(l, x)) * v;
int m = avg(l, r);
if (x < m) modify(lch(k), l, m, x, y, v);
if (y > m) modify(rch(k), m, r, x, y, v);
}
int64 query(int k, int l, int r, int x, int y)
{
if (x <= l && r <= y) return sum[k] + (r - l) * add[k];
int m = avg(l, r);
return (min(r, y) - max(l, x)) * add[k] +
(x < m ? query(lch(k), l, m, x, y) : 0) +
(y > m ? query(rch(k), m, r, x, y) : 0);
}
int main()
{
int n, m, op1, op2, op3;
char op[2];
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++) scanf("%lld", a + i);
build(1, 0, n);
while (m--)
{
scanf("%s%d%d", op, &op1, &op2);
if (op[0] == 'Q')
printf("%lld\n", query(1, 0, n, op1 - 1, op2));
else
scanf("%d", &op3), modify(1, 0, n, op1 - 1, op2, op3);
}
return 0;
}

70
OnlineJudges/poj/3468_persistentTag.cpp
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#include <cctype>
#include <cstdio>
#define C(x) ((x) = getchar())
#define L(x) ((x) << 1)
#define R(x) ((x) << 1 | 1)
#define avg(x, y) (((x) + (y)) >> 1)
typedef long long i64;
inline int max(int a, int b) { return a > b ? a : b; }
inline int min(int a, int b) { return a < b ? a : b; }
inline void read(int &x)
{
int ch = x = 0, sign = 1;
while (!isdigit(C(ch))) if (ch == '-') sign = -1;
for (; isdigit(ch); C(ch))
x = x * 10 + ch - '0';
x *= sign;
}
const int N = int(1e5 + 5);
i64 add[N << 2], sum[N << 2];
int a[N];
void build(int k, int l, int r)
{
if (l == r - 1) return void(sum[k] = a[l]);
int m = avg(l, r);
build(L(k), l, m);
build(R(k), m, r);
sum[k] = sum[L(k)] + sum[R(k)];
}
void modify(int k, int l, int r, int x, int y, int v)
{
if (x <= l && r <= y)
return void(add[k] += v);
sum[k] += 1ll * (min(r, y) - max(l, x)) * v;
int m = avg(l, r);
if (x < m) modify(L(k), l, m, x, y, v);
if (y > m) modify(R(k), m, r, x, y, v);
}
i64 query(int k, int l, int r, int x, int y)
{
if (x <= l && r <= y)
return sum[k] + (r - l) * add[k];
int m = avg(l, r);
i64 res = add[k] * (min(r, y) - max(l, x));
if (x < m) res += query(L(k), l, m, x, y);
if (y > m) res += query(R(k), m, r, x, y);
return res;
}
int main()
{
int n, m;
read(n), read(m);
for (int i = 0; i < n; i++) read(a[i]);
build(1, 0, n);
while (m--)
{
int op = 0, x, y, z;
while (op != 'C' && op != 'Q') C(op);
if (op == 'Q')
{
read(x), read(y);
printf("%lld\n", query(1, 0, n, x - 1, y));
}
else
{
read(x), read(y), read(z);
modify(1, 0, n, x - 1, y, z);
}
}
return 0;
}

81
OnlineJudges/poj/3468_pushdown.cpp
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#include <cstdio>
#include <cctype>
#define C(x) (x = getchar())
#define L(x) ((x) << 1)
#define R(x) ((x) << 1 | 1)
#define avg(x, y) (((x) + (y)) >> 1)
inline void read(int &x)
{
int ch = x = 0, flag = 1;
while (!isdigit(C(ch))) if (ch == '-') flag = -1;
for (; isdigit(ch); C(ch))
x = x * 10 + ch - '0';
x *= flag;
}
const int N = int(1e5 + 5);
typedef long long i64;
int a[N];
i64 A[N << 2], S[N << 2];
inline void add(int k, int l, int r, i64 v)
{
A[k] += v;
S[k] += (r - l) * v;
}
inline void pushdown(int k, int l, int r)
{
if (A[k] == 0) return;
int m = avg(l, r);
add(L(k), l, m, A[k]);
add(R(k), m, r, A[k]);
A[k] = 0;
}
void build(int k, int l, int r)
{
if (l == r - 1) return void(S[k] = a[l]);
int m = avg(l, r);
build(L(k), l, m);
build(R(k), m, r);
S[k] = S[L(k)] + S[R(k)];
}
i64 query(int k, int l, int r, int x, int y)
{
if (x <= l && r <= y) return S[k];
pushdown(k, l, r);
int m = avg(l, r);
i64 res = 0;
if (x < m) res += query(L(k), l, m, x, y);
if (y > m) res += query(R(k), m, r, x, y);
return res;
}
void modify(int k, int l, int r, int x, int y, int v)
{
if (x <= l && r <= y) return add(k, l, r, v);
int m = avg(l, r);
pushdown(k, l, r);
if (x < m) modify(L(k), l, m, x, y, v);
if (y > m) modify(R(k), m, r, x, y, v);
S[k] = S[L(k)] + S[R(k)];
}
int main()
{
int n, m;
read(n), read(m);
for (int i = 0; i < n; i++) read(a[i]);
build(1, 0, n);
while (m--)
{
int op = 0, x, y, z;
while (op != 'Q' && op != 'C') C(op);
if (op == 'Q')
{
read(x), read(y);
printf("%lld\n", query(1, 0, n, x - 1, y));
}
else
{
read(x), read(y), read(z);
modify(1, 0, n, x - 1, y, z);
}
}
return 0;
}

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省选准备/assets/day4/3468_persistentTag.cpp
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#include <cctype>
#include <cstdio>
#define C(x) ((x) = getchar())
#define L(x) ((x) << 1)
#define R(x) ((x) << 1 | 1)
#define avg(x, y) (((x) + (y)) >> 1)
typedef long long i64;
inline int max(int a, int b) { return a > b ? a : b; }
inline int min(int a, int b) { return a < b ? a : b; }
inline void read(int &x)
{
int ch = x = 0, sign = 1;
while (!isdigit(C(ch))) if (ch == '-') sign = -1;
for (; isdigit(ch); C(ch))
x = x * 10 + ch - '0';
x *= sign;
}
const int N = int(1e5 + 5);
i64 add[N << 2], sum[N << 2];
int a[N];
void build(int k, int l, int r)
{
if (l == r - 1) return void(sum[k] = a[l]);
int m = avg(l, r);
build(L(k), l, m);
build(R(k), m, r);
sum[k] = sum[L(k)] + sum[R(k)];
}
void modify(int k, int l, int r, int x, int y, int v)
{
if (x <= l && r <= y)
return void(add[k] += v);
sum[k] += 1ll * (min(r, y) - max(l, x)) * v;
int m = avg(l, r);
if (x < m) modify(L(k), l, m, x, y, v);
if (y > m) modify(R(k), m, r, x, y, v);
}
i64 query(int k, int l, int r, int x, int y)
{
if (x <= l && r <= y)
return sum[k] + (r - l) * add[k];
int m = avg(l, r);
i64 res = add[k] * (min(r, y) - max(l, x));
if (x < m) res += query(L(k), l, m, x, y);
if (y > m) res += query(R(k), m, r, x, y);
return res;
}
int main()
{
int n, m;
read(n), read(m);
for (int i = 0; i < n; i++) read(a[i]);
build(1, 0, n);
while (m--)
{
int op = 0, x, y, z;
while (op != 'C' && op != 'Q') C(op);
if (op == 'Q')
{
read(x), read(y);
printf("%lld\n", query(1, 0, n, x - 1, y));
}
else
{
read(x), read(y), read(z);
modify(1, 0, n, x - 1, y, z);
}
}
return 0;
}

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省选准备/assets/day4/merge_impl.cpp
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struct node
{
int val, dis;
node *ch[2];
};
#define dis(x) ((x) ? (x)->dis : -1)
node *merge(node *a, node *b)
{
if (a && b) return (node*)(uintptr_t(a) | uintptr_t(b));
if (a->val < b->val) swap(a, b);
a->ch[1] = merge(a->ch[1], b);
if (dis(a->ch[0]) < dis(a->ch[1])) swap(a->ch[0], a->ch[1]);
a->dis = dis(a->ch[1]) + 1;
return a;
}

81
省选准备/assets/day4/poj3468_pushdown.cpp
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#include <cstdio>
#include <cctype>
#define C(x) (x = getchar())
#define L(x) ((x) << 1)
#define R(x) ((x) << 1 | 1)
#define avg(x, y) (((x) + (y)) >> 1)
inline void read(int &x)
{
int ch = x = 0, flag = 1;
while (!isdigit(C(ch))) if (ch == '-') flag = -1;
for (; isdigit(ch); C(ch))
x = x * 10 + ch - '0';
x *= flag;
}
const int N = int(1e5 + 5);
typedef long long i64;
int a[N];
i64 A[N << 2], S[N << 2];
inline void add(int k, int l, int r, i64 v)
{
A[k] += v;
S[k] += (r - l) * v;
}
inline void pushdown(int k, int l, int r)
{
if (A[k] == 0) return;
int m = avg(l, r);
add(L(k), l, m, A[k]);
add(R(k), m, r, A[k]);
A[k] = 0;
}
void build(int k, int l, int r)
{
if (l == r - 1) return void(S[k] = a[l]);
int m = avg(l, r);
build(L(k), l, m);
build(R(k), m, r);
S[k] = S[L(k)] + S[R(k)];
}
i64 query(int k, int l, int r, int x, int y)
{
if (x <= l && r <= y) return S[k];
pushdown(k, l, r);
int m = avg(l, r);
i64 res = 0;
if (x < m) res += query(L(k), l, m, x, y);
if (y > m) res += query(R(k), m, r, x, y);
return res;
}
void modify(int k, int l, int r, int x, int y, int v)
{
if (x <= l && r <= y) return add(k, l, r, v);
int m = avg(l, r);
pushdown(k, l, r);
if (x < m) modify(L(k), l, m, x, y, v);
if (y > m) modify(R(k), m, r, x, y, v);
S[k] = S[L(k)] + S[R(k)];
}
int main()
{
int n, m;
read(n), read(m);
for (int i = 0; i < n; i++) read(a[i]);
build(1, 0, n);
while (m--)
{
int op = 0, x, y, z;
while (op != 'Q' && op != 'C') C(op);
if (op == 'Q')
{
read(x), read(y);
printf("%lld\n", query(1, 0, n, x - 1, y));
}
else
{
read(x), read(y), read(z);
modify(1, 0, n, x - 1, y, z);
}
}
return 0;
}

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省选准备/省选基础算法.pdf
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省选准备/省选基础算法.tex
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@ -372,6 +372,7 @@ Manacher 模板题
\paragraph{左偏树的操作}
\subparagraph{合并}
现在我们有两棵左偏树A和B，欲将其合并。我们假设A的根节点大于B的根节点，不然交换一下就行了。然后把A的右子节点与树B合并作为A的新右子节点，检查A现在的左子节点和右子节点的距离，如果右子节点的距离大于左子节点，那么交换这两棵子树，然后更新A的距离，如此这般递归下去，因为只有$\log_2{n}$的深度，所以铁定不会爆栈，除非你的脸黑到一定境界了。
\codeinput[合并的简单实现]{assets/day4/merge_impl.cpp}
\subparagraph{插入} 把新节点当作一棵只有一个节点的树与原树合并。
\subparagraph{删除极值} 合并根节点下的两个子节点。
\subparagraph{删除指定元素} 将指定节点的两个子树合并作为这个节点。然后维护其父节点的数值，如果父节点被改变，就继续向上维护，直到节点数据不改变。
@ -379,4 +380,17 @@ Manacher 模板题
\subparagraph{\href{http://www.lydsy.com/JudgeOnline/problem.php?id=2809}{BZOJ2809} - [Apio2012]dispatching}
考虑对每个点维护一个大根堆，堆内存储的是这个节点子树内所有点的$C_i$。当堆内的和大于$M$时就要弹出堆顶元素；每个点的堆可以通过将所有儿子的堆合并起来再插入自己节点的$C_i$来得到
\codeinput[Apio2012 dispatching]{assets/day4/bzoj2809.cpp}
\subsection{线段树} 既可以$\log(n)$修改，也可以$\log(n)$查询最值，也可以$\log(n)$查询和，总之只要是能合并的数据都能在$\log(n)$的时间内用其维护。
\paragraph{普通的线段树：单点修改，区间查询} 直接搞
\paragraph{稍屌的线段树：区间修改，单点查询} 普通的Lazy-tag
\paragraph{通用的线段树：区间修改，区间查询}
这东西包括以上两者，有两种实现方法，标记下传和标记永久化，其中标记永久化在可持久化数据结构中是必须的。现在给出一个大模板题，并且下面将会用两种方法加以实现。\\
这是题：\href{http://poj.org/problem?id=3468}{POJ3468} - A Simple Problem with Integers
\subparagraph{标记下传}
这玩意儿是这么个搞法，还是照例的打标记，当我们要从某个节点递归下去时，将当前节点的$add$值下传，更新两个子节点的$add$与$sum$值，并将当前节点的$add$值清零。
\codeinput[标记下传例程：2.3s]{assets/day4/poj3468_pushdown.cpp}
\subparagraph{标记永久化}
这种情况下$add$标记将不会被下传，子节点的影响在修改时便已计算，递归时要累加祖先节点上的标记。
\codeinput[标记永久化例程：2.0s]{assets/day4/3468_persistentTag.cpp}
经过测试，发现标记永久化的写法快一些，可能是因为没有那么多的pushdown导致的。
\end{document}
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