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:

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

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\n",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;
}

转载于:https://www.cnblogs.com/shenben/p/6270672.html

POJ 2887 Big String相关推荐

oracle rman备份时间点,Oracle Rman 控制RMAN的备份时间，减少IO消耗
一.问题描述由于服务器配置不高,备份策略为周末全备.周一至周六差异备份. 平时服务器CPU使用30%左右. 全备份时,开启两个通道,CPU达到70%-80%左右,业务不卡顿.不掉单,session不 ...
Poj 2887-Big String Splay
题目:http://poj.org/problem?id=2887 Big String Time Limit: 1000MS Memory Limit: 131072K Total Submis ...
POJ 1002 487-3279
2019独角兽企业重金招聘Python工程师标准>>> 487-3279 Time Limit: 2000MS Memory Limit: 65536K Total Submissi ...
字符串匹配のKMP【专题@AbandonZHANG】
算法详解很长时间内都没有能够很理解KMP算法的精髓,尤其是很多书上包括<算法导论>没有把next函数(亦或 π函数)讲解的很透彻. 今天去看了matrix67大牛博客中关于kmp部分的讲 ...
字符串匹配のKMP【@Abandon】
算法详解很长时间内都没有能够很理解KMP算法的精髓,尤其是很多书上包括<算法导论>没有把next函数(亦或 π函数)讲解的很透彻. 今天去看了matrix67大牛博客中关于kmp部分的讲 ...
Java知识——精华总结
Java知识--精华总结一.java概述与基础知识 1.何为编程? 编程就是让计算机为解决某个问题而使用某种程序设计语言编写程序代码,并最终得到结果的过程. 为了使计算机能够理解人的意图,人类就必须 ...
【贪心】Sunscreen（poj 3614/luogu 2887）
Sunscreen poj 3614 luogu 2887 题目大意: 有n个人,每个人要求选一个价值在minniminn_iminni到maxximaxx_imaxxi的物品,现在有m件物品,每 ...
POJ 3268 D-Silver Cow Party
http://poj.org/problem?id=3268 Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently ...
POJ 2226 Muddy Fields 最小点覆盖+加建图（好题）
题目链接题目一看就是最小点覆盖,这道题与POJ - 3041 算是一类题,但是3041算是一道十分裸的,因为删除的是整行或者整列,所以图其实是现成的,但是本题的难点就在如何建图. 思路:首先还是尽量 ...

POJ 2887 Big String

题目大意

POJ 2887 Big String相关推荐

最新文章

热门文章