BZOJ 4999 This Problem Is Too Simple!

维护区间里面有几个 $x$ 的最好方法

BZOJ 4999 This Problem Is Too Simple!

题目描述

给你一棵树,需要支持以下操作:

  • 1 x y :将 $A_x$ 修改为 $y$
  • 2 x y p :查询从 $x$ 到 $y$ 的路径上有多少个结点等于 $p$

题解

线段树动态开点,对于每个数建一棵权值线段树,写法类似于主席树

AC Code

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//#define NOSTDCPP
//#define Cpp11
//#define Linux_System
#ifndef NOSTDCPP
#include <bits/stdc++.h>
#else
#include <algorithm>
#include <bitset>
#include <cassert>
#include <climits>
#include <complex>
#include <cstring>
#include <cstdio>
#include <deque>
#include <exception>
#include <functional>
#include <iomanip>
#include <iostream>
#include <istream>
#include <iterator>
#include <list>
#include <map>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>
#endif
# ifdef Linux_System
# define getchar getchar_unlocked
# define putchar putchar_unlocked
# endif
# define RESET(_) memset(_, 0, sizeof(_))
# define RESET_(_, val) memset(_, val, sizeof(_))
# define fi first
# define se second
# define pb push_back
# define midf(x, y) ((x + y) >> 1)
# define DXA(_) ((_ << 1))
# define DXB(_) ((_ << 1) | 1)
# define next __Chtholly__
# define x1 __Mercury__
# define y1 __bbtl04__
# define index __ikooo__
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector <int> vi;
typedef set <int> si;
typedef pair <int, int> pii;
typedef long double ld;
const int MOD = 1e9 + 7;
const int maxn = 100009;
const int maxm = 300009;
const ll inf = 1e18;
const double pi = acos(-1.0);
const double eps = 1e-6;
ll myrand(ll mod){return ((ll)rand() << 32 ^ (ll)rand() << 16 ^ rand()) % mod;}
template <class T>
inline bool scan_d(T & ret)
{
char c;
int sgn;
if(c = getchar(), c == EOF)return false;
while(c != '-' && (c < '0' || c > '9'))c = getchar();
sgn = (c == '-') ? -1 : 1;
ret = (c == '-') ? 0 : (c - '0');
while(c = getchar(), c >= '0' && c <= '9')
ret = ret * 10 + (c - '0');
ret *= sgn;
return true;
}
#ifdef Cpp11
template <class T, class ... Args>
inline bool scan_d(T & ret, Args & ... args)
{
scan_d(ret);
scan_d(args...);
}
#define cin.tie(0); cin.tie(nullptr);
#define cout.tie(0); cout.tie(nullptr);
#endif
inline bool scan_ch(char &ch)
{
if(ch = getchar(), ch == EOF)return false;
while(ch == ' ' || ch == '\n')ch = getchar();
return true;
}
template <class T>
inline void out_number(T x)
{
if(x < 0)
{
putchar('-');
out_number(- x);
return ;
}
if(x > 9)out_number(x / 10);
putchar(x % 10 + '0');
}
int n, m;
struct edge
{
int to, next;
}edg[maxm];
int fir[maxn], nedge;
inline void addedge(int f, int t)
{
edg[++ nedge].to = t;
edg[nedge].next = fir[f];
fir[f] = nedge;
}
int deep[maxn], father[maxn], son[maxn], size[maxn], head[maxn], id[maxn], cnt_HLD;
void dfs1(int u, int fa, int d)
{
deep[u] = d;
father[u] = fa;
son[u] = 0;
size[u] = 1;
for(int i = fir[u]; i; i = edg[i].next)
{
int v = edg[i].to;
if(v != fa)
{
dfs1(v, u, d + 1);
size[u] += size[v];
if(! son[u] || size[v] > size[son[u]])
son[u] = v;
}
}
}
void getpos(int u, int sp)
{
head[u] = sp;
id[u] = ++ cnt_HLD;
if(son[u]) getpos(son[u], sp);
for(int i = fir[u]; i; i = edg[i].next)
{
int v = edg[i].to;
if(v != son[u] && v != father[u])
getpos(v, v);
}
}
inline void init_HLD()
{
cnt_HLD = 0;
RESET(fir);
RESET(son);
RESET(father);
RESET_(deep, -1);
RESET(size);
RESET(head);
}
struct node
{
int l, r, sum;
}tree[maxm * 50];
int CMT_cnt, root[maxm];
inline void init_CMT()
{
CMT_cnt = 0;
RESET(root);
}
int l, r, v;
void update(int nl, int nr, int &p)
{
if(! p) p = ++ CMT_cnt;
tree[p].sum += v;
if(nl == nr) return ;
int mid = midf(nl, nr);
if(l <= mid) update(nl, mid, tree[p].l);
else update(mid + 1, nr, tree[p].r);
}
int query(int nl, int nr, int p)
{
if(! p) return 0;
if(l <= nl && nr <= r) return tree[p].sum;
int mid = midf(nl, nr), ans = 0;
if(l <= mid) ans += query(nl, mid, tree[p].l);
if(mid < r) ans += query(mid + 1, nr, tree[p].r);
return ans;
}
inline int __ans__(int u, int v, int d)
{
int f1 = head[u], f2 = head[v], ans = 0;
while(f1 != f2)
{
if(deep[f1] < deep[f2])
{
swap(f1,f2);
swap(u,v);
}
l = id[f1], r = id[u];
ans += query(1, n, root[d]);
u = father[f1]; f1 = head[u];
}
if(deep[u] > deep[v]) swap(u, v);
l = id[u], r = id[v];
ans += query(1, n, root[d]);
return ans;
}
map <int, int> mp;
int A[maxn], MAP_cnt;
int main()
{
int f, t, d;
char ch;
scanf("%d %d", &n, &m);
MAP_cnt = 0;
for(int i = 1; i <= n; ++ i)
{
scanf("%d", A + i);
if(! mp[A[i]]) mp[A[i]] = ++ MAP_cnt;
}
for(int i = 1; i < n; ++ i)
{
scanf("%d %d", &f, &t);
addedge(f, t);
addedge(t, f);
}
dfs1(1, -1, 0);
getpos(1, 1);
v = 1;
for(int i = 1; i <= n; ++ i)
{
l = id[i];
update(1, n, root[mp[A[i]]]);
}
while(m --)
{
scanf("\n%c %d %d", &ch, &f, &t);
switch(ch)
{
case 'C':
if(! mp[t]) mp[t] = ++ MAP_cnt;
l = id[f];
v = -1;
update(1, n, root[mp[A[f]]]);
v = 1;
update(1, n, root[mp[t]]);
A[f] = t;
break;
case 'Q':
scanf("%d", &d);
if(! mp[d]) puts("0");
else printf("%d\n", __ans__(f, t, mp[d]));
break;
default:
assert(ch == 'C' || ch == 'Q');
}
//if(CMT_cnt >= maxm){while(true); puts("GG");}
}
return 0;
}
/*
5 6
10 20 30 40 50
1 2
1 3
3 4
3 5
Q 2 3 40
C 1 40
Q 2 3 40
Q 4 5 30
C 3 10
Q 4 5 30
*/
文章目录
  1. 1. BZOJ 4999 This Problem Is Too Simple!
    1. 1.1. 题目描述
    2. 1.2. 题解
    3. 1.3. AC Code
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