Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals) C. String Reconstruction 並查集

C. String Reconstruction

題目鏈接：

http://codeforces.com/contest/828/problem/C

Description

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.

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 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.

Output

Print lexicographically minimal string that fits all the information Ivan remembers.

Sample Input

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

Sample Output

abacaba

Hint

題意

給n個字符串，告訴你有n個位置的是這個字符串的開始，而後讓你輸出最後字符串的樣子。

題解：

因爲顯然每一個位置的字符是惟一的，換句話而言，就是每一個位置咱們都只用訪問一次就夠了，不必重複的進行訪問。

咱們用並查集去維護這個就行了。

代碼

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
const int maxn = 2e6+7;
int fa[maxn];
int ans[maxn];
int fi(int x){
    return x==fa[x]?x:fa[x]=fi(fa[x]);
}
int n,mx=0;
string s;
int main(){
    scanf("%d",&n);
    for(int i=0;i<maxn;i++)fa[i]=i;
    int mx = 0;
    for(int i=0;i<n;i++){
        cin>>s;
        int k;
        scanf("%d",&k);
        for(int j=0;j<k;j++){
            int x;
            scanf("%d",&x);
            int st=x;
            int end=x+s.size();
            mx=max(mx,end);
            x=fi(x);
            while(x<end){
                ans[x]=s[x-st]-'a';
                int x2=fi(fi(x)+1);
                fa[fi(x)]=x2;
                x=x2;
            }
        }
    }
    for(int i=1;i<mx;i++)
        cout<<char(ans[i]+'a');
    cout<<endl;
}