【BZOJ 2453】 维护队列

【题目链接】

https://www.lydsy.com/JudgeOnline/problem.php?id=2453

【算法】

带修改的莫队算法

当块的大小为N^(2/3)时，时间复杂度为 : O(n^(5/3))

但本题，当块的大小为sqrt(N)时，能够达到更优的复杂度

【代码】

#include<bits/stdc++.h>
using namespace std;
#define MAXN 10010
const int MAXC = 1e6 + ;

int i,len,x,y,sum,l,r,n,now,qnum,m,cnt;
int a[MAXN],block[MAXN],s[MAXC],ans[MAXN];
char op[];

template <typename T> inline void read(T &x)
{
    int f = ; x = ;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
    for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
    x *= f;
}
template <typename T> inline void write(T x)
{
    if (x < )
    {
        putchar('-');
        x = -x;
    }
    if (x > ) write(x/);
    putchar(x%+'');
}
template <typename T> inline void writeln(T x)
{
    write(x);
    puts("");
}
struct info
{
    int x,y,last;
} c[MAXN];
struct Query
{
    int x,y,id,cur;
} q[MAXN];

inline bool cmp(Query a,Query b)
{
    if (block[a.x] == block[b.x])
    {
        if (block[a.y] == block[b.y]) return a.cur < b.cur;
        else return block[a.y] < block[b.y];
    } else return block[a.x] < block[b.x];
}
inline void add(int x)
{
    s[a[x]]++;
    if (s[a[x]] == ) sum++;
}
inline void dec(int x)
{
    s[a[x]]--;
    if (s[a[x]] == ) sum--;
}
inline void worka(int now)
{
    int x = c[now].x,y = c[now].y;
    if (x >= l && x <= r) dec(x);
    c[now].last = a[x];
    a[x] = y;
    if (x >= l && x <= r) add(x);
}
inline void workb(int now)
{
    int x = c[now].x,y = c[now].y;
    if (x >= l && x <= r) dec(x);
    a[x] = c[now].last;
    if (x >= l && x <= r) add(x);
}

int main()
{

    read(n); read(m);
    len = sqrt(n);
    for (i = ; i <= n; i++) block[i] = (i - ) / len + ;
    for (i = ; i <= n; i++) read(a[i]);
    for (i = ; i <= m; i++)
    {
        scanf("%s",&op);
        read(x); read(y);
        if (op[] == 'Q') q[++qnum] = (Query){x,y,qnum,cnt};
        else c[++cnt] = (info){x,y,};
    }
    sort(q+,q+qnum+,cmp);
    l = ; r = ; now = ; sum = ;
    for (i = ; i <= qnum; i++)
    {
        for (; r < q[i].y; r++) add(r+);
        for (; r > q[i].y; r--) dec(r);
        for (; l < q[i].x; l++) dec(l);
        for (; l > q[i].x; l--) add(l-);
        for (now; now < q[i].cur; now++) worka(now+);
        for (now; now > q[i].cur; now--) workb(now);
        ans[q[i].id] = sum;
    }
    for (i = ; i <= qnum; i++) printf("%d\n",ans[i]);

    return ;
}