【JZOJ 4594】【UVa 12345】Dynamic len

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Description

有n个数编号从0→n-1,两种操作：
Q L R:询问编号为L→R-1的数中共有多少种不同的数
M X Y：将编号为X的数改为Y
共有m个操作

Solution

此题用莫队算法，
很裸嘛~,直接上带修改即可，
复杂度：O(n53)

Code

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<algorithm>#define fo(i,a,b) for(int i=a;i<=b;i++)#define fod(i,a,b) for(int i=a;i>=b;i--)#define asp1(q) {if(!z[q])ans++;z[q]++;}#define asj1(q) {z[q]--;if(!z[q])ans--;}using namespace std;const int N=1000500;int read(int &n){    char ch=' ';int q=0,w=1;    for(;(ch!='-')&&((ch<'0')||(ch>'9'));ch=getchar());    if(ch=='-')w=-1,ch=getchar();    for(;ch>='0' && ch<='9';ch=getchar())q=q*10+ch-48;n=q*w;return n;}int m1,m,n,ans;int a[N],s[N];int z[N];struct qqww{int x,y,z,i,la;}b[N],b1[N];bool PX(qqww q,qqww w){    int x1=(q.x-1)/m1+1,x2=(w.x-1)/m1+1;    int y1=ceil(1.0*q.y/m1),y2=ceil(1.0*w.y/m1);    return(x1<x2||(x1==x2&&y1<y2)||(x1==x2&&y1==y2&&q.z<w.z)||(x1==x2&&y1==y2&&q.z==w.z&&q.x<w.x));}int main(){    freopen("len.in","r",stdin);    freopen("len.out","w",stdout);    int q,w,e;    read(n);read(m);    m1=pow(n,1.0/3);    m1*=m1;     fo(i,1,n)z[i]=read(a[i]);    e=0;    fo(i,1,m)    {        char ch=' ';while(ch!='Q'&&ch!='M')ch=getchar();        read(q),read(w);q++;        if(ch=='M')b1[++e].x=q,b1[e].y=w;            else b[i-e].x=q,b[i-e].y=w,b[i-e].z=e,b[i-e].i=i-e;    }    m-=e;    sort(b+1,b+1+m,PX);    fo(i,1,e)b1[i].la=z[b1[i].x],z[b1[i].x]=b1[i].y;    memset(z,0,sizeof(z));    int l=b[1].x,r=b[1].x-1;ans=0;    fo(i,1,m)    {        fo(j,b[i-1].z+1,b[i].z)        {            if(l<=b1[j].x&&b1[j].x<=r)            {                asj1(a[b1[j].x]);                asp1(b1[j].y);            }            a[b1[j].x]=b1[j].y;        }        fod(j,b[i-1].z,b[i].z+1)        {            if(l<=b1[j].x&&b1[j].x<=r)            {                asj1(a[b1[j].x]);                asp1(b1[j].la);            }            a[b1[j].x]=b1[j].la;        }        fo(j,r+1,b[i].y)asp1(a[j]);        fod(j,r,b[i].y+1)asj1(a[j]);        fo(j,l,b[i].x-1)asj1(a[j]);        fod(j,l-1,b[i].x)asp1(a[j]);        s[b[i].i]=ans;        l=b[i].x,r=b[i].y;    }    fo(i,1,m)printf("%d\n",s[i]);    return 0;}

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