poj 2887 块状数组/线段树

Big String

Time Limit: 1000MS Memory Limit: 131072KTotal Submissions: 5952 Accepted: 1405

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

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

Sample Output

ade

Source

POJ Monthly--2006.07.30, zhucheng

字符串总长度不超过10^6，插入操作不超过2000次。可以用块状数组存储字符串，比如10^6分成1000行存，每行1000个，并用len[]数组记录每行的字符串长度。插入第k个就找到第k个字符应该属于哪一行，直接插入，操作次数也不超过改行字符串长度也就是几千。

块状数组思想是平方分割法，将长度为n的数组分sqrt(n)块，每个长度为n/sqrt(n)=sqrt(n)，对每个区间进行操作只需要O(sqrt(n))的复杂度。

这题是插入和查询第k个点，一般还可以用线段树离线输出结果。先把所有的查询和插入都进来，倒叙模拟插入，用线段树更新和统计区间和。

#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;#define maxn 1001char s[maxn*maxn];char a[maxn][maxn*3]; //插入最多2000次，每行最多3000个int len[maxn];  //记录每一行字符个数const int b = 1000; //默认分成最多1000行int row; //共几行char query(int k) //查询第k个{    int cnt = 0;    for(int i = 0; i < row; i++){        if(cnt+len[i]>= k){            return a[i][k-cnt-1];        }        cnt+=len[i];    }}void add(char ch, int k) //插入到第k个{    int cnt = 0;    int r;    for(int i = 0; i < row; i++){        if(cnt+len[i] >= k){            r = i;            break;        }        if(i == row-1) {r = row-1; break;}        cnt+=len[i];    }    int pos = k-cnt-1;    if(pos >= len[r]) pos = len[r];    for(int i = len[r]; i >= pos+1; i--)        a[r][i] = a[r][i-1];    a[r][pos] = ch;    len[r]++;}int main(){    while(scanf("%s", s)==1){             memset(len, 0, sizeof(len));            int slen = strlen(s);            int perl = (slen+b-1)/b; //每行保存的字符个数            row = (slen-1)/perl+1; //共多少行            for(int i = 0; i < row; i++) len[i] = perl;            for(int i = 0; i < slen; i++)                a[i/perl][i%perl] = s[i];            len[row-1] = (slen-1)%perl+1;            int n;            scanf("%d", &n);            while(n--){                char op[3];                int k;                scanf("%s", op);                if(op[0] == 'Q'){                    scanf("%d", &k);                    printf("%c\n", query(k));                }                else{                    scanf("%s %d", op, &k);                    add(op[0], k);                }            }    }    return 0;}