poj 2887 Big String

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Big String

Time Limit: 1000MS Memory Limit: 131072KTotal Submissions: 5792 Accepted: 1353

Description

You are given a string and supposed to do some string manipulations.

Input

The first line of the input contains the initial string. You can assume that it is non-empty and its length does not exceed 1,000,000.

The second line contains the number of manipulation commands N (0 < N ≤ 2,000). The following N lines describe a command each. The commands are in one of the two formats below:

I ch p: Insert a character ch before the p-th character of the current string. If p is larger than the length of the string, the character is appended to the end of the string.
Q p: Query the p-th character of the current string. The input ensures that the p-th character exists.

All characters in the input are digits or lowercase letters of the English alphabet.

Output

For each Q command output one line containing only the single character queried.

Sample Input

ab7Q 1I c 2I d 4I e 2Q 5I f 1Q 3

Sample Output

ade

这题是用块链过的，差点被卡时间了，接近700ms。因为我是写的动态分配型的，不会写数组版的。

但这题如果用其他数据结构，貌似会TLE，因为O(nlogn)的建树时间，在n=10^6的情况下，那是伤不起的！

但是说句实话，块链真不适合竞赛写！

代码：

#include<cstdio>#include<iostream>#include<cstring>#include<cstdlib>#define Maxn 1000010using namespace std;char s[Maxn];const int max_size=1000;struct blocklist{    char num[max_size];    int sz;    blocklist* next;    blocklist():sz(0),next(0){}}*head=NULL;blocklist* get_pos(int &x){    blocklist* ret=head;    while(ret&&x>ret->sz){        x-=ret->sz;        ret=ret->next;    }    if(!ret) x=1;    return ret;}bool merge(blocklist* pos){    if(!pos||!pos->next) return false;    if(pos->sz+pos->next->sz<=max_size){        memmove(pos->num+pos->sz,pos->next->num,sizeof(char)*pos->next->sz);        pos->sz+=pos->next->sz;        blocklist* u=pos->next;        pos->next=u->next;        delete u;        return true;    }    return false;}void maintain(){    blocklist* pos=head;    while(pos){        merge(pos);        pos=pos->next;    }}void split(blocklist* pos,int x){    blocklist* tmp=pos->next;    pos->next=new blocklist();    pos->next->next=tmp;    memmove(pos->next->num,pos->num+x-1,sizeof(char)*(pos->sz-x+1));    pos->next->sz=pos->sz-x+1;    pos->sz=x-1;}void insert(int x,int n,char* ch){    blocklist* pos=get_pos(x);    if(pos) split(pos,x);    else{        pos=new blocklist();        if(!head) head=pos;        else{            blocklist* it;            for(it=head;it->next;it=it->next);            it->next=pos;        }    }    for(int i=0;i<n;i++){        if(pos->sz==max_size){            blocklist* tmp=pos->next;            pos->next=new blocklist();            pos->next->next=tmp;            pos=pos->next;        }        pos->num[pos->sz++]=ch[i];    }    maintain();}char query(int n){    blocklist* pos=head;    while(n>pos->sz){        n-=pos->sz;        pos=pos->next;    }    return pos->num[n-1];}int main(){    char ch[10];    int n,p,x;    scanf("%s",s);    int len=strlen(s);    insert(1,len,s);    scanf("%d",&n);    for(int i=0;i<n;i++){        scanf("%s",ch);        if(ch[0]=='Q'){            scanf("%d",&x);            printf("%c\n",query(x));        }        else{            scanf("%s%d",ch,&p);            insert(p,1,ch);        }    }return 0;}

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