Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals) C. String Reconstruction 并查集
C. String Reconstruction
题目连接:
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
题意
给n个字符串,告诉你有n个位置的是这个字符串的开始,然后让你输出最后字符串的样子。
题解:
由于显然每个位置的字符是唯一的,换句话而言,就是每个位置我们都只用访问一次就够了,没必要重复的进行访问。
我们用并查集去维护这个就好了。
代码
#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;
}
Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals) C. String Reconstruction 并查集的更多相关文章
- 【贪心】Codeforces Round #423 (Div. 1, rated, based on VK Cup Finals) A. String Reconstruction
在每个给出的子串的起始位置打个标记,记录的是从这里开始的最长子串. 然后输出的时候就扫,如果遇到开始位置,就从这里开始输出,如果其后被更长的覆盖,就跳转到更长的串进行输出. 如果位置没被覆盖,就输出' ...
- Codeforces Round #423 (Div. 1, rated, based on VK Cup Finals)
Codeforces Round #423 (Div. 1, rated, based on VK Cup Finals) A.String Reconstruction B. High Load C ...
- Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals) E. DNA Evolution 树状数组
E. DNA Evolution 题目连接: http://codeforces.com/contest/828/problem/E Description Everyone knows that D ...
- Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals) Problem E (Codeforces 828E) - 分块
Everyone knows that DNA strands consist of nucleotides. There are four types of nucleotides: "A ...
- Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals) Problem C (Codeforces 828C) - 链表 - 并查集
Ivan had string s consisting of small English letters. However, his friend Julia decided to make fun ...
- Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals) D. High Load 构造
D. High Load 题目连接: http://codeforces.com/contest/828/problem/D Description Arkady needs your help ag ...
- Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals) A,B,C
A.题目链接:http://codeforces.com/contest/828/problem/A 解题思路: 直接暴力模拟 #include<bits/stdc++.h> using ...
- Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals) Problem D (Codeforces 828D) - 贪心
Arkady needs your help again! This time he decided to build his own high-speed Internet exchange poi ...
- Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals) Problem A - B
Pronlem A In a small restaurant there are a tables for one person and b tables for two persons. It i ...
随机推荐
- python--使用递归优雅实现列表相加和进制转换
咦,好像坚持了一段时间,感觉又有新收获啦. # coding: utf-8 class Stack: def __init__(self): self.items = [] # 是否为空 def is ...
- Newtonsoft.Json WindowPhone7.1
才发现最新版本的Newtonsoft.Json 已经不支持wp71了… 查了下最后一个支持的版本是Json.NET 5.0 Release 8… 安装的时候需要: Install-Package Ne ...
- 最小生成树模板【kruskal & prim】
CDOJ 1966 Kruskal 解法 时间复杂度O(mlogm) m为边数,这里主要是边排序占时间,后面并查集还好 #include <cstdio> #include <cst ...
- angular 2+ 变化检测系列一(基础概念)
什么是变化检测? 变化检测的基本功能就是获取应用程序的内部状态(state),并且是将这种状态对用户界面保持可见.状态可以是javascript中的任何的数据结构,比如对象,数组,(数字,布尔,字符串 ...
- 清北合肥day1
题目: 1.给出一个由0,1组成的环 求最少多少次交换(任意两个位置)使得0,1靠在一起 n<=1000 2.两个数列,支持在第一个数列上区间+1,-1 每次花费为1 求a变成b的最小代价 n& ...
- grails2.3.3发布了-【grails】
grails2.3.3发布了,在grails2.3.2中默认方式fork模式下无法运行的BUG也解决了. 需要做的相关修改为修改BuildConfig.groovy: build ':tomcat:7 ...
- arcgis 10 File GDB转成9.3 File GDB
背景:已有的File GDB数据是在ArcGIS 10下面创建的,在运行9.3的电脑上无法读取,需要转换成低版本. 方法: 1.在ArcGIS Toolbox中找到相关工具(datamanagemen ...
- git之二: git可视化工具sourcetree
参考: https://www.cnblogs.com/tian-xie/p/6264104.html sourcetree安装使用
- window下php5.5安装redis扩展
redis是现在比较流行的noSQL,主流大型网站都用的比较多,很多同学不知道怎么安装,这里介绍在windows下面安装以及扩展,提供学习使用,实际使用环境多在Linux下. 1.phpinfo(), ...
- 51Nod1577 异或凑数 线性基
原文链接https://www.cnblogs.com/zhouzhendong/p/51Nod1577.html 题意 给定一个长度为 n 的序列. 有 m 组询问,每一组询问给出 L,R,k ,询 ...