Codeforces 828C String Reconstruction【思维+并查集】

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本文介绍了一种算法，用于根据给定的子字符串及其在目标字符串中的位置信息，构造出符合要求的最小字典序字符串。该问题通过使用并查集优化处理，确保了高效的求解过程。

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 t_i occurs in string s at least k_i times or more, he also remembers exactly k_i positions where the string t_i occurs in string s: these positions are x_i, 1, x_i, 2, ..., x_{i, k_i}. He remembers n such strings t_i.

You are to reconstruct lexicographically minimal string s such that it fits all the information Ivan remembers. Strings t_i and string s consist of small English letters only.

Input

The 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 t_i, then positive integer k_i, which equal to the number of times the string t_i occurs in string s, and then k_i distinct positive integers x_i, 1, x_i, 2, ..., x_{i, k_i} in increasing order — positions, in which occurrences of the string t_i in the string s start. It is guaranteed that the sum of lengths of strings t_i doesn't exceed 10⁶, 1 ≤ x_i, j ≤ 10⁶, 1 ≤ k_i ≤ 10⁶, and the sum of all k_i doesn't exceed 10⁶. The strings t_i can coincide.

It is guaranteed that the input data is not self-contradictory, and thus at least one answer always exists.

Output

Print 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

题目大意：

现在让你构造出来一个最小字典序的字符串，使得其满足n个条件。

这n个条件中每个条件有一个连续的子字符串。

然后有Ki个位子，表示这些字符串出现的位子。

保证有解，n个条件互相没有矛盾。

思路：

问题其实就是一个涂色问题。每次在Ki个点处作为起点，然后向后涂这个子字符串，如果已经涂过的位子不用再涂，否则涂上这个字母。

那么暴力去涂色肯定是要超时的。

我们维护一个并查集，使得find（x）查找到的位子是位子x之后，第一个没有涂色过的位子。

那么我们在暴力涂色的过程中，可以用并查集优化寻找速度。

很古老的套路题了= =

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int f[3000000];
char ans[3000000];
char a[3000000];
int find(int a)
{
    int r=a;
    while(f[r]!=r)
    r=f[r];
    int i=a;
    int j;
    while(i!=r)
    {
        j=f[i];
        f[i]=r;
        i=j;
    }
    return r;
}
void merge(int a,int b)
{
    int A,B;
    A=find(a);
    B=find(b);
    if(A>B)
    {
        f[B]=A;
    }
    else f[A]=B;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(ans,' ',sizeof(ans));
        int end=0;
        for(int i=1;i<=3000000-80;i++)f[i]=i;
        for(int i=1;i<=n;i++)
        {
            scanf("%s",a);
            int lenn=strlen(a);
            int ki;
            scanf("%d",&ki);
            while(ki--)
            {
                int x;
                scanf("%d",&x);
                int tmpx=x;
                end=max(end,lenn+x-1);
                while(x<=tmpx+lenn-1)
                {
                    ans[x]=a[x-tmpx+1-1];
                    x=find(x);
                    if(x<=tmpx+lenn-1)merge(x,x+1);
                }
            }
        }
        for(int i=1;i<=end;i++)
        {
            if(ans[i]==' ')ans[i]='a';
            printf("%c",ans[i]);
        }
        printf("\n");
    }
}