Codeforces 221d D. Little Elephant and Array
二次联通门 : Codeforces 221d D. Little Elephant and Array
/*
Codeforces 221d D. Little Elephant and Array 题意 : 询问一段区间中出现次数等于自身的数的个数 正解是dp 莫队水过, 作为我莫队的入门题
myj的思路 66 把所有需查询的区间排序 当前查询区间的答案为上一个区间的答案通过多次的区间移动得出 */
#include <algorithm>
#include <cstdio>
#include <cmath> #define Max 100005 void read (int &now)
{
now = ;
register char word = getchar ();
bool temp = false;
while (word < '' || word > '')
{
if (word == '-')
temp = true;
word = getchar ();
}
while (word >= '' && word <= '')
{
now = now * + word - '';
word = getchar ();
}
if (temp)
now = -now;
} int N, M; int belong[Max];
struct Data
{
int l, r; int ID; bool operator < (const Data &now) const
{
if (belong[now.l] == belong[this->l])
return now.r > this->r;
return belong[now.l] > belong[this->l];
}
}; int number[Max]; int K_Size;
int count[Max];
int rank_[Max]; Data query[Max];
int Answer[Max], Result;
int pos[Max]; incount void Updata (int now, int key)
{
if (count[number[now]] == pos[now])
Result--;
count[number[now]] += key;
if (count[number[now]] == pos[now])
Result++;
} int main (int argc, char *argv[])
{
read (N);
read (M);
K_Size = sqrt (N);
for (int i = ; i <= N; i++)
{
read (number[i]);
rank_[i] = number[i];
pos[i] = number[i];
belong[i] = (i + ) / K_Size;
}
std :: sort (rank_ + , rank_ + + N);
int Size = std :: unique (rank_ + , rank_ + + N) - rank_ - ;
for (int i = ; i <= N; i++)
number[i] = std :: lower_bound (rank_ + , rank_ + + Size, number[i]) - rank_;
for (int i = ; i <= M; i++)
{
read (query[i].l);
read (query[i].r);
query[i].ID = i;
}
std :: sort (query + , query + + M);
int l = , r = ;
for (int cur = , now_1, now_2; cur <= M; cur++)
{
now_1 = query[cur].l;
now_2 = query[cur].r;
if (l < now_1)
for (int i = l; i < now_1; i++)
Updata (i, -);
else
for (int i = l - ; i >= now_1; i--)
Updata (i, );
if (r < now_2)
for (int i = r + ; i <= now_2; i++)
Updata (i, );
else
for (int i = r; i > now_2; i--)
Updata (i, -);
l = now_1;
r = now_2;
Answer[query[cur].ID] = Result;
}
for (int i = ; i <= M; i++)
printf ("%d\n", Answer[i]);
return ;
}
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