POJ 2887 Big String

Big String

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 7053		Accepted: 1684

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:

1. 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.
2. 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

Source

POJ Monthly--2006.07.30, zhucheng

题目大意 

给一个字符串，长度不超过 10⁶，有两种操作：

    1、Q x(0 < x <= len(s)) 表示查询当前串中第x个字符

    2、I c x(c为字母 0 < x <= len(s)+1)表示在第x个位置插入c字符 x == len+1表示在串尾插入

操作的总数不超过 2000

做法分析

    块状链表裸题。详见代码

#include<cstdio>
#include<cstring>
#include<iostream>
#define m(s) memset(s,0,sizeof s)
using namespace std;
const int N=1010;
int l[N],n,m;
char s[N*N],eg[N][N*3];
void insert(int x,char c){
    int n1=0,p1,pn=n;
    for(int i=1;i<=n;i++){
        if(n1+l[i]>=x){pn=i;break;}
        if(i==n) break;
        n1+=l[i];
    }
    p1=x-n1;l[pn]=max(p1,l[pn]+1);
    for(int i=l[pn];i>p1;i--) eg[pn][i]=eg[pn][i-1];
    eg[pn][p1]=c;
}
void query(int x){
    int n1=0,p1,pn=n;
    for(int i=1;i<=n;i++){
        if(n1+l[i]>=x){pn=i;break;}
        n1+=l[i];
    }
    p1=x-n1;
    printf("%c
",eg[pn][p1]);
}
void work(){
    scanf("%d",&m);
    int len=strlen(s),ave;
    ave=(len+999)/1000;
    n=(len-1)/ave+1;
    for(int i=0;i<len;i++) eg[i/ave+1][i%ave+1]=s[i],l[i/ave+1]++;
    while(m--){
        char c[2];int x;
        scanf("%s",c);
        if(c[0]=='I') scanf("%s%d",c,&x),insert(x,c[0]);
        else scanf("%d",&x),query(x);
    }
}
int main(){
    while(scanf("%s",s)==1){
        m(l);m(eg);work();
    } 
    return 0;
}