String Reconstruction

最新推荐文章于 2022-03-19 21:00:00 发布

原创最新推荐文章于 2022-03-19 21:00:00 发布 · 185 阅读

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本文探讨了基于特定条件重构字符串的问题，包括字符串的重复次数和位置信息，旨在寻找字典序最小的解决方案。通过算法设计，文章提供了实现这一目标的代码实例。

Ivan had string s consisting of small English letters. However, his friend Julia decided to make fun of him and hid the string s. Ivan preferred making a new string to finding the old one.
Ivan knows some information about the string s. Namely, he remembers, that string ti occurs in string s at least ki times or more, he also remembers exactly ki positions where the string ti occurs in string s: these positions are xi, 1, xi, 2, …, xi, ki. He remembers n such strings ti.
You are to reconstruct lexicographically minimal string s such that it fits all the information Ivan remembers. Strings ti and string s consist of small English letters only.InputThe first line contains single integer n (1 ≤ n ≤ 105) — the number of strings Ivan remembers.
The next n lines contain information about the strings. The i-th of these lines contains non-empty string ti, then positive integer ki, which equal to the number of times the string ti occurs in string s, and then ki distinct positive integers xi, 1, xi, 2, …, xi, ki in increasing order — positions, in which occurrences of the string ti in the string s start. It is guaranteed that the sum of lengths of strings ti doesn’t exceed 106, 1 ≤ xi, j ≤ 106, 1 ≤ ki ≤ 106, and the sum of all ki doesn’t exceed 106. The strings ti can coincide.
It is guaranteed that the input data is not self-contradictory, and thus at least one answer always exists.OutputPrint lexicographically minimal string that fits all the information Ivan remembers. Examples

Input

3
a 4 1 3 5 7
ab 2 1 5
ca 1 4

Output

abacaba

Input

1
a 1 3

Output

aaa

Input

3
ab 1 1
aba 1 3
ab 2 3 5

Output

ababab

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <iostream>
using namespace std;
char a[2000006],b[2000006];
#define inf 0x3f3f3f3f
int main()
{
 int n,ans=0;
 cin>>n;
 while(n--)
 {
  int t=-99999,k,m;
  cin>>b>>m;
  int l=strlen(b);
  while(m--)
  {
   cin>>k;
   ans=max(ans,k+l-1);
   for(int i=max(t,k-1);i<(l+k-1);i++)
   {
    a[i]=b[i-k+1];
   }
   t=l+k-1;
  }
  } 
  for(int i=0;i<ans;i++)
  {
   if(a[i]=='\0')
   printf("a");
   else
   printf("%c",a[i]);
  }
  printf("\n");
}