poj 2887 Big String

Big String

Time Limit: 1000MS		Memory Limit: 131072K
Total 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

ab
7
Q 1
I c 2
I d 4
I e 2
Q 5
I f 1
Q 3

Sample Output

a
d
e

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 1000010
using 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;
}