POJ2887 分块

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Big String

Time Limit: 1000MS Memory Limit: 131072KTotal Submissions: 6670 Accepted: 1583

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

Source

POJ Monthly--2006.07.30, zhucheng

题意：给一个字符串，长度<=1000000;P(P<=2000)个操作,向位置添加字符或询问该处字符；

据说可以线段树，但是我不会写，就写了个分块；

分成sqrt(n+p)块，每次二分找到要对哪一块进行操作，暴力O(sqrt(n+p))插入，O(1)查找,因为sqrt(n+p)在1000左右，所以不会超时

代码：

#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<iostream>#include<cmath>const int maxn = 1000010;using namespace std;char s[maxn];struct block{int size;char c[1010];void insert(int pos,char x){for(int i=++size;i>pos;i--) c[i]=c[i-1];c[pos]=x;}char query(int pos){return c[pos];}void push_back(char x){c[++size]=x;}}a[1010];int p;int len;int cnt;int sum[1010];int find(int left,int right,int x){int l=left,r=right;int ans=0;while(l<=r){int mid=(l+r)>>1;if(sum[mid]>=x){ans=mid;r=mid-1;}else   l=mid+1;}return ans;}void getnum(int r){for(int i=r;i<=cnt;i++)   sum[i]=sum[i-1]+a[i].size;}void init(){len=strlen(s)+p;cnt=sqrt(len*1.0)+1;for(int i=0;i<len;i++)   a[i/cnt+1].push_back(s[i]);getnum(1);}void inc(char x,int pos){int p=find(1,cnt,pos);a[p].insert(pos-sum[p-1],x);getnum(p);}char search(int pos){int p=find(1,cnt,pos);return a[p].query(pos-sum[p-1]);}int main(){scanf("%s",s);scanf("%d",&p);init();while(p--){char f[10];int x;char ss[10];scanf("%s",f);if(f[0]=='I'){scanf("%s%d",ss,&x);inc(ss[0],x);}else{scanf("%d",&x);printf("%c\n",search(x));}}return 0;}

~~感觉自己水平越来越差了，，，为什么铺天盖地都是数据结构题！！！~~

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