HDU 5933

奇怪的莫队搞法

HDU 5933 ArcSoft’s Office Rearrangement

题目描述

原来有 $N$ 堆数 现在要分成 $M$ 堆都相等的

每次只能相邻的合并,或者一堆分成多堆

问最少要多少次操作

题解

将多的向少的分配即可。刚好的就跳过了。

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, 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. HDU 5933 ArcSoft’s Office Rearrangement
    1. 1.1. 题目描述
    2. 1.2. 题解
    3. 1.3. AC Code
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