XDOJ 1393

线段树维护树的点

XDOJ 1393 海未的剑道

题目描述

询问一个点的子树里面有多少个点的深度小于 $k$

题解

对这棵树求括号序,对括号序利用线段树维护深度最大值和深度最小值:

  • 对于询问,如果区间所有的点的深度在题意范围内,直接返回整个区间的大小;否则继续递归,直到深度都不在题意范围内,结束递归。

AC Code

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
//#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, unit;
struct query
{
int l, r, id;
bool operator < (const query &b) const
{
return l / unit < b.l / unit || (l / unit == b.l / unit && r < b.r);
}
}qqq[maxn];
int ans[maxn], num[maxn], data[maxn];
void modui()
{
ll temp = 0;
RESET(num);
int l = 0, r = n + 1;
for(register int i = 1; i <= m; ++ i)
{
while(r < qqq[i].r)
{
temp -= num[data[r]] == 1;
-- num[data[r]];
++ r;
}
while(r > qqq[i].r)
{
-- r;
++ num[data[r]];
temp += num[data[r]] == 1;
}
while(l > qqq[i].l)
{
temp -= num[data[l]] == 1;
-- num[data[l]];
-- l;
}
while(l < qqq[i].l)
{
++ l;
++ num[data[l]];
temp += num[data[l]] == 1;
}
ans[qqq[i].id] = temp;
}
}
int main()
{
ios :: sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
while(~ scanf("%d %d", &n, &m))
{
unit = sqrt(n);
for(register int i = 1; i <= n; ++ i) scanf("%d", data + i);
for(register int i = 1; i <= m; ++ i)
{
scanf("%d %d", &qqq[i].l, &qqq[i].r);
qqq[i].id = i;
}
sort(qqq + 1, qqq + 1 + m);
modui();
for(register int i = 1; i <= m; ++ i)
printf("%d\n", ans[i]);
}
return 0;
}
文章目录
  1. 1. XDOJ 1393 海未的剑道
    1. 1.1. 题目描述
    2. 1.2. 题解
    3. 1.3. AC Code
{{ live2d() }}