CF-796B
B. Find The Bonetime limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.
InputThe 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.
OutputPrint one integer — the final position along the x-axis of the bone.
Examplesinput7 3 4
3 4 6
1 2
2 5
5 7
7 1output1input5 1 2
2
1 2
2 4output2NoteIn 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.
题意:
有n个杯子,m个杯子底下有洞,将骨头交换k次,遇到洞就会掉下去不动,求最后骨头的位置。
注意交换的两个位置是不分先后的。
AC代码:
#include<bits/stdc++.h>
using namespace std; int main(){
ios::sync_with_stdio(false);
int h[];
int n,m,k,t,s,flag=,ans=;
cin>>n>>m>>k;
//scanf("%d%d%d",&n,&m,&k);
memset(h,,sizeof(h));
for(int i=;i<m;i++){
cin>>t;
//scanf("%d",&t);
h[t]=;
}
if(h[]){
flag=;
}
for(int i=;i<k;i++){
cin>>t>>s;
//scanf("%d%d",&t,&s);
if(!h[t]&&ans==t){
ans=s;
}
else if(!h[s]&&ans==s){
ans=t;
}
}
cout<<ans<<endl;
return ;
}
CF-796B的更多相关文章
- ORA-00494: enqueue [CF] held for too long (more than 900 seconds) by 'inst 1, osid 5166'
凌晨收到同事电话,反馈应用程序访问Oracle数据库时报错,当时现场现象确认: 1. 应用程序访问不了数据库,使用SQL Developer测试发现访问不了数据库.报ORA-12570 TNS:pac ...
- cf之路,1,Codeforces Round #345 (Div. 2)
cf之路,1,Codeforces Round #345 (Div. 2) ps:昨天第一次参加cf比赛,比赛之前为了熟悉下cf比赛题目的难度.所以做了round#345连试试水的深浅..... ...
- cf Round 613
A.Peter and Snow Blower(计算几何) 给定一个点和一个多边形,求出这个多边形绕这个点旋转一圈后形成的面积.保证这个点不在多边形内. 画个图能明白 这个图形是一个圆环,那么就是这个 ...
- ARC下OC对象和CF对象之间的桥接(bridge)
在开发iOS应用程序时我们有时会用到Core Foundation对象简称CF,例如Core Graphics.Core Text,并且我们可能需要将CF对象和OC对象进行互相转化,我们知道,ARC环 ...
- [Recommendation System] 推荐系统之协同过滤(CF)算法详解和实现
1 集体智慧和协同过滤 1.1 什么是集体智慧(社会计算)? 集体智慧 (Collective Intelligence) 并不是 Web2.0 时代特有的,只是在 Web2.0 时代,大家在 Web ...
- CF memsql Start[c]UP 2.0 A
CF memsql Start[c]UP 2.0 A A. Golden System time limit per test 1 second memory limit per test 256 m ...
- CF memsql Start[c]UP 2.0 B
CF memsql Start[c]UP 2.0 B B. Distributed Join time limit per test 1 second memory limit per test 25 ...
- CF #376 (Div. 2) C. dfs
1.CF #376 (Div. 2) C. Socks dfs 2.题意:给袜子上色,使n天左右脚袜子都同样颜色. 3.总结:一开始用链表存图,一直TLE test 6 (1)如果需 ...
- CF #375 (Div. 2) D. bfs
1.CF #375 (Div. 2) D. Lakes in Berland 2.总结:麻烦的bfs,但其实很水.. 3.题意:n*m的陆地与水泽,水泽在边界表示连通海洋.最后要剩k个湖,总要填掉多 ...
- CF #374 (Div. 2) D. 贪心,优先队列或set
1.CF #374 (Div. 2) D. Maxim and Array 2.总结:按绝对值最小贪心下去即可 3.题意:对n个数进行+x或-x的k次操作,要使操作之后的n个数乘积最小. (1)优 ...
随机推荐
- Arduino关于旋转编码器程序的介绍(Reading Rotary Encoders)--by Markdown
介绍 旋转或编码器是一个角度測量装置. 他用作精确測量电机的旋转角度或者用来控制控制轮子(能够无限旋转,而电位器只能旋转到特定位置).其中有一些还安装了一个能够在轴上按的button,就像音乐播放器的 ...
- C#读取自定义的config
今天说下C#读写自定义config文件的各种方法.由于这类文章已经很多,但是大多数人举例子都是默认的在app.confg或者web.config进行读写,而不是一般的XML文件,我主要写的是一般的Xm ...
- JAVA 数据筛选(第一笔数据与第二笔数据比较)
第一笔数据与第二笔数据比较 Map<String, Object> jHpictureMap = new HashMap<String, Object>(); // 存放照片S ...
- 游戏开发之基础图像---3d图像处理
http://dev.gameres.com/Program/Visual/3D/3Darit.htm float 类型数据有效数字是小数点后面6位 单精度 doluble 类型数据是至少10位,双精 ...
- 从 Spring Cloud 看一个微服务框架的「五脏六腑」(转)
Spring Cloud 是一个基于 Spring Boot 实现的微服务框架,它包含了实现微服务架构所需的各种组件. 本文将从 Spring Cloud 出发,分两小节讲述微服务框架的「五脏六腑」: ...
- PHP开发工作心得
一.扎实PHP自身的基础知识.函数.常量等,尽量用内置的方法解决这个问题(由于个人写的往往运行效率没有内置方法高): 二.代码尽量少的实现功能(由于PHP的运行事实上是,将咱们的代码先处理成底层语言进 ...
- mysql分页查询-limit
分页查询的sql: select * from table limit 4,10; 4表示查询的索引,索引是从0开始,4表示从第五条数据开始查询,10表示要查询多少条数据,10表示查询十条数据 如果从 ...
- ABAP DEMO-2018
sap Program DEMO 介绍 Program Description BALVBT01 Example SAP program for displying multiple ALV repo ...
- guava cache与spring集成
缓存的背景 缓存,在我们日常开发中是必不可少的一种解决性能问题的方法.简单的说,cache 就是为了提升系统性能而开辟的一块内存空间.在cpu进行计算的时候, 首先是读取寄存器,然后内存,再是硬盘.由 ...
- iOS反射机制:objc_property_t的使用
#import <objc/runtime.h> 需要导入这个头文件. 动态获取一个自定义类对象中的所有属性 - (NSDictionary *)allProperties { NSMut ...
