BZOJ 3289

令人卡常的操作

简单的主席树,当然树状数组也行

BZOJ 3289 Mato的文件管理

题目描述

给一些数,询问你 $[l, r]$ 里面有多少个逆序对

题解

树状数组 + 莫队。

莫队对查询排序,树状数组求逆序对。

注意是动态求逆序对

具体实现看代码

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 = 50009;
const int maxm = 65539;
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, blk, A[maxn];
ll ans[maxn];
ll tree[maxm];
inline int lowbit(int x){return x & -x;}
inline void update(int x, int val)
{
for(int i = x; i < maxn; i += lowbit(i))
tree[i] += val;
}
inline ll sum(int x)
{
if(! x) return 0;
ll ans = 0;
for(int i = x; i; i -= lowbit(i))
ans += tree[i];
return ans;
}
struct query
{
int l, r, id;
query(int l = 1, int r = 1, int id = 1) : l(l), r(r), id(id) {}
bool operator < (const query &b) const
{
return l / blk == b.l / blk ? r < b.r : l / blk < b.l / blk;
}
}qqq[maxn];
vi g;
inline void modui()
{
ll tmp = 0;
int L = 1, R = 0;
for(int i = 1; i <= m; ++ i)
{
while(L < qqq[i].l)
{
update(A[L], -1);
tmp -= sum(A[L] - 1);
++ L;
}
while(L > qqq[i].l)
{
-- L;
update(A[L], 1);
tmp += sum(A[L] - 1);
}
while(R < qqq[i].r)
{
++ R;
update(A[R], 1);
tmp += R - L + 1 - sum(A[R]);
}
while(R > qqq[i].r)
{
update(A[R], -1);
tmp -= R - L - sum(A[R]);
-- R;
}
ans[qqq[i].id] = tmp;
}
}
int main()
{
g.clear();
scanf("%d", &n);
for(register int i = 1; i <= n; ++ i) scanf("%d", A + i), g.push_back(A[i]);
sort(g.begin(), g.end());
g.erase(unique(g.begin(), g.end()), g.end());
for(register int i = 1; i <= n; ++ i)
A[i] = lower_bound(g.begin(), g.end(), A[i]) - g.begin() + 1;
scanf("%d", &m);
blk = (int)sqrt(1.0 * m);
for(int i = 1, f, t; i <= m; ++ i)
{
scanf("%d %d", &f, &t);
qqq[i] = query(f, t, i);
}
sort(qqq + 1, qqq + 1 + m);
modui();
for(int i = 1; i <= m; ++ i)
printf("%lld\n", ans[i]);
return 0;
}
文章目录
  1. 1. BZOJ 3289 Mato的文件管理
    1. 1.1. 题目描述
    2. 1.2. 题解
    3. 1.3. AC Code
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