LOJ #6284. Getting started with series partitioning 8,

LOJ #6284. Getting started with series partitioning 8,
#6284. sequence partitioning entry 8 memory limit: 256 MiB time limit: 500 ms standard input and output question type: traditional evaluation method: text comparison uploaded by: hzwer submit Record statistics discussion 1 Test Data question description

A series with a length of nnn and an nnn operation are provided. The operation involves the query of an element that is equal to a number ccc, and all the elements in this range are changed to ccc.

Input Format

Enter a number nnn in the first line.

Enter nnn numbers in the second row. the I-th digit is aia_iai, separated by spaces.

Next, enter the nnn line and enter three numbers lll, rrr, and ccc in each line, separated by spaces.

It indicates that the number in [l, r] [l, r] [l, r] Is ccc, and then the number in [l, r] [l, r, the numbers of r] [l, r] are changed to ccc.

Output Format

For each query, a row of numbers is output to indicate the answer.

Sample Input

41 2 2 41 3 11 4 41 2 21 4 2

Sample output

1102

Data range and prompt

For data of 100% 100 \ % 100%, 1 ≤ n ≤ 100000, − 231 ≤ others 1 \ leq n \ leq 100000, -2 ^ {31} \ leq \ mathrm {others} 1 ≤ n ≤ 100000, − 231 ≤ others, ans ≤ 231 − 1 \ mathrm {ans} \ leq 2 ^ {31}-1ans ≤ 231 − 1.

Classification label hzwer question set partitioning and data structure by size Maintain the value of each block. Fragmented chunks are marked as violent Proof of time complexity:

 

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;const int MAXN=1e5+10,mod=10007;inline char nc(){    static char buf[MAXN],*p1=buf,*p2=buf;    return p1==p2&&(p2=(p1=buf)+fread(buf,1,MAXN,stdin)),p1==p2?EOF:*p1++;}inline int read(){    char c=nc();int x=0,f=1;    while(c<'0'||c>'9'){if(c=='-')f=-1;c=nc();}    while(c>='0'&&c<='9'){x=x*10+c-'0';c=nc();}    return x*f;}int a[MAXN],belong[MAXN],L[MAXN],R[MAXN],block,tag[MAXN];void reset(int x){    if(tag[x]==-1) return ;    for(int i=L[x*block];i<=R[x*block];i++)        a[i]=tag[x];    tag[x]=-1;}int Query(int l,int r,int val){    int ans=0;    reset(belong[l]);    for(int i=l;i<=min(r,R[l]);i++)    {        if(a[i]==val) ans++;        a[i]=val;    }    if(belong[l]!=belong[r])    {        reset(belong[r]);        for(int i=L[r];i<=r;i++)        {            if(a[i]==val) ans++;            a[i]=val;        }    }    for(int i=belong[l]+1;i<=belong[r]-1;i++)    {        if(tag[i]==-1)        {            for(int j=L[i*block];j<=R[i*block];j++)                if(a[j]==val)                    ans++;                else a[j]=val;        }        else if(tag[i]==val) ans+=block;        tag[i]=val;    }    return ans;}int main(){    #ifdef WIN32    freopen("a.in","r",stdin);    #else    #endif    int N=read();block=sqrt(N);    memset(tag,-1,sizeof(tag));    for(int i=1;i<=N;i++) a[i]=read(),belong[i]=(i-1)/block+1,L[i]=(belong[i]-1)*block+1,R[i]=belong[i]*block;    for(int i=1;i<=N;i++)    {        int l=read(),r=read(),c=read();        printf("%d\n",Query(l,r,c));    }    return 0;}