CodeForces - 796B-Find The Bone(模拟)
Zane the wizard is going to perform a magic show shuffling the cups.
There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x = i.
The problematic bone is initially at the position x = 1. Zane will confuse the audience by swapping the cups k times, the i-th time of which involves the cups at the positions x = ui and x = vi. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations.
Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at x = 4 and the one at x = 6, they will not be at the position x = 5 at any moment during the operation.
Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone.
Input
The first line contains three integers n, m, and k (2 ≤ n ≤ 106, 1 ≤ m ≤ n, 1 ≤ k ≤ 3·105) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively.
The second line contains m distinct integers h1, h2, ..., hm (1 ≤ hi ≤ n) — the positions along the x-axis where there is a hole on the table.
Each of the next k lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the positions of the cups to be swapped.
Output
Print one integer — the final position along the x-axis of the bone.
Examples
Input
7 3 4
3 4 6
1 2
2 5
5 7
7 1
Output
1
Input
5 1 2
2
1 2
2 4
Output
2
Note
In the first sample, after the operations, the bone becomes at x = 2, x = 5, x = 7, and x = 1, respectively.
In the second sample, after the first operation, the bone becomes at x = 2, and falls into the hole onto the ground.
题解:模拟来回交换的过程即可
代码
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int vis[2000005];
int main() {
int n,m,k;
cin>>n>>m>>k;
int s;
for(int t=0; t<m; t++) {
scanf("%d",&s);
vis[s]=1;
}
int a,b,temp=1;
int cnt,flag=0;
for(int t=0; t<k; t++) {
scanf("%d%d",&a,&b);
if(a==temp&&vis[temp]==0) {
temp=b;
} else if(b==temp&&vis[temp]==0) {
temp=a;
}
}
cout<<temp<<endl;
return 0;
}
:
CodeForces - 796B-Find The Bone(模拟)的更多相关文章
- CodeForces.158A Next Round (水模拟)
CodeForces.158A Next Round (水模拟) 题意分析 校赛水题的英文版,坑点就是要求为正数. 代码总览 #include <iostream> #include &l ...
- 【codeforces 796B】Find The Bone
[题目链接]:http://codeforces.com/contest/796/problem/B [题意] 一开始骨头在1号位置; 然后有m个洞,给出洞的下标; 然后有k个交换操作; 如果骨头到洞 ...
- CodeForces - 796B 模拟
思路:模拟移动即可,如果球落入洞中停止移动.注意:有可能第一个位置就是洞!! AC代码 #include <cstdio> #include <cmath> #include ...
- Codeforces 747C:Servers(模拟)
http://codeforces.com/problemset/problem/747/C 题意:有n台机器,q个操作.每次操作从ti时间开始,需要ki台机器,花费di的时间.每次选择机器从小到大开 ...
- Codeforces 740A. Alyona and copybooks 模拟
A. Alyona and copybooks time limit per test: 1 second memory limit per test: 256 megabytes input: st ...
- Codeforces 716A Crazy Computer 【模拟】 (Codeforces Round #372 (Div. 2))
A. Crazy Computer time limit per test 2 seconds memory limit per test 256 megabytes input standard i ...
- CodeForces 670 A. Holidays(模拟)
Description On the planet Mars a year lasts exactly n days (there are no leap years on Mars). But Ma ...
- Codeforces 280D k-Maximum Subsequence Sum [模拟费用流,线段树]
洛谷 Codeforces bzoj1,bzoj2 这可真是一道n倍经验题呢-- 思路 我首先想到了DP,然后矩阵,然后线段树,然后T飞-- 搜了题解之后发现是模拟费用流. 直接维护选k个子段时的最优 ...
- Codeforces 1090B - LaTeX Expert - [字符串模拟][2018-2019 Russia Open High School Programming Contest Problem B]
题目链接:https://codeforces.com/contest/1090/problem/B Examplesstandard input The most famous characters ...
- 796B Find The Bone
B. Find The Bone time limit per test 2 seconds memory limit per test 256 megabytes input standard in ...
随机推荐
- jsp九大内置对象和其作用详解
jsp九大内置对象和其作用详解 JSP中一共预先定义了9个这样的对象,分别为:request.response.session.application.out.pagecontext.config.p ...
- Speaking 1
What clothes do you usually like to wear?Well I like fashionable clothes, but I also want to be comf ...
- julia
版本还不成熟,等成熟了再完整的看看吧.
- GSM —— 商业蜂窝通信系统
用户漫游: HLR:当用户办卡时,当地运营商把用户资料(归属地信息等)输入 HLR: VLR:当用户漫游到别的城市时,漫游地的 VLR 把用户资料从 HLR 复制过来,用户便可以继续享受运营商的通信服 ...
- BZOJ1798:[AHOI2009]维护序列
浅谈树状数组与线段树:https://www.cnblogs.com/AKMer/p/9946944.html 题目传送门:https://www.lydsy.com/JudgeOnline/prob ...
- zoom在清除浮动中的利用
zoom 是个困惑了好久的元素,今天对它有了个初步的认识 zoom , ie 的专属属性,在其他浏览器中不起作用,它的原本功能是设置或检测对象的缩放比例(只在ie下起作用) 比如 <div ...
- Algorithms & Data structures in C++& GO ( Lock Free Queue)
https://github.com/xtaci/algorithms //已实现 ( Implemented ): Array shuffle https://github.com/xtaci/al ...
- 【247】◀▶IEW-Unit12
Unit 12 Leisure Activities 1.Model1对应图片分析 2.Model1范文分析 The pie chart shows the six sporting activiti ...
- App Distribution Guide--(三)---Configuring Your Xcode Project for Distribution
Configuring Your Xcode Project for Distribution You can edit your project settings anytime, but some ...
- Linux rpm 命令参数使用…
RPM是RedHat Package Manager(RedHat软件包管理工具)类似Windows里面的"添加/删除程序" rpm 执行安装包 二进制包(Binary)以及源代码 ...