PTA 05-树8 File Transfer (25分)
题目地址
https://pta.patest.cn/pta/test/16/exam/4/question/670
5-8 File Transfer (25分)
We have a network of computers and a list of bi-directional connections. Each of these connections allows a file transfer from one computer to another. Is it possible to send a file from any computer on the network to any other?
Input Specification:
Each input file contains one test case. For each test case, the first line contains NN (2\le N\le 10^42≤N≤104), the total number of computers in a network. Each computer in the network is then represented by a positive integer between 1 and NN. Then in the following lines, the input is given in the format:
I c1 c2
where I stands for inputting a connection between c1 and c2; or
C c1 c2
where C stands for checking if it is possible to transfer files between c1 and c2; or
S
where S stands for stopping this case.
Output Specification:
For each C case, print in one line the word "yes" or "no" if it is possible or impossible to transfer files between c1 and c2, respectively. At the end of each case, print in one line "The network is connected." if there is a path between any pair of computers; or "There are k components." where k is the number of connected components in this network.
Sample Input 1:
5
C 3 2
I 3 2
C 1 5
I 4 5
I 2 4
C 3 5
S
Sample Output 1:
no
no
yes
There are 2 components.
Sample Input 2:
5
C 3 2
I 3 2
C 1 5
I 4 5
I 2 4
C 3 5
I 1 3
C 1 5
S
Sample Output 2:
no
no
yes
yes
The network is connected.
这题是考并查集
/*
运行详情
评测结果
时间 结果 得分 题目 编译器 用时(ms) 内存(MB) 用户
2017-06-29 18:12 部分正确 24 5-8 gcc 27 1
测试点结果
测试点 结果 得分/满分 用时(ms) 内存(MB)
测试点1 答案正确 7/7 18 1
测试点2 答案正确 7/7 1 1
测试点3 答案正确 1/1 2 1
测试点4 答案正确 1/1 2 1
测试点5 答案正确 4/4 14 1
测试点6 答案正确 4/4 15 1
测试点7 答案错误 0/1 27 1 没有ac,不知道最后一个测试点用了什么数据
*/ #include<stdio.h>
#define MAXLEN 15000
#define TOPLEVEL -1
int gComputers[MAXLEN];
void InitComputers(int n)
{
int i;
for(i=1;i<=n;i++)
{
gComputers[i]=TOPLEVEL;
}
}
void swap(int *a,int *b)
{
int temp;
temp=*a;
*a=*b;
*b=temp;
}
void Connect(int a,int b)
{
/* if(gComputers[a]==TOPLEVEL)
{
gComputers[b]=a;
return;
}
*/
int temp;
temp=a;
while(gComputers[temp]!=TOPLEVEL)
{
temp=gComputers[temp];
}
// gComputers[a]=temp;
gComputers[b]=temp;
return;
} /*
int GetTop(int a)
{ int temp=a;
if(gComputers[a]==TOPLEVEL)
return a; while(gComputers[temp]!=TOPLEVEL)
temp=gComputers[temp];
gComputers[a]=temp; //查找完了合并父节点;
return temp; }
*/
int GetTop(int k)
{
if(gComputers[k]<0)
{
return k;
}
return gComputers[k] = GetTop(gComputers[k]);
} void Check(int a,int b)
{
if(GetTop(a)==GetTop(b))
{
printf("yes\n");
}
else
{
printf("no\n");
}
} void ShowSummry(int n)
{
int i,count=0;
for(i=1;i<=n;i++)//此处扫描应从1号开始,电脑从1开始编号,0号没操作始终是-1;
{
if(gComputers[i]==TOPLEVEL)
{
count++;
}
}
if(count==1)
{
printf("The network is connected.");
}
else
printf("There are %d components.",count);
} void OutputStatus(int n)
{
int i;
for(i=0;i<=n;i++)
{
printf("--%d:%d\n",i,gComputers[i]);
}
} int main()
{
int N;
int a,b;
char command;
scanf("%d",&N);
InitComputers(N);
while(1)
{
scanf("%c",&command);
if(command=='S')
{
ShowSummry(N);
// OutputStatus(N);
break;
} scanf("%d %d",&a,&b);
if(command=='I')
Connect(a,b);
if(command=='C')
Check(a,b);
}
}
不知道最后一个测试点什么情况,mooc开课之外的题库里面不带提示,找了一张别人做题的截图,发现最后一个点是卡不压缩的,容易超时
问题是我有一直合并路线,而且时间也不长。不知道遇上什么问题了。

PTA 05-树8 File Transfer (25分)的更多相关文章
- PAT 5-8 File Transfer (25分)
We have a network of computers and a list of bi-directional connections. Each of these connections a ...
- 05-树8 File Transfer (25 分)
We have a network of computers and a list of bi-directional connections. Each of these connections a ...
- PTA甲级1094 The Largest Generation (25分)
PTA甲级1094 The Largest Generation (25分) A family hierarchy is usually presented by a pedigree tree wh ...
- L2-006 树的遍历 (25 分) (根据后序遍历与中序遍历建二叉树)
题目链接:https://pintia.cn/problem-sets/994805046380707840/problems/994805069361299456 L2-006 树的遍历 (25 分 ...
- pat04-树5. File Transfer (25)
04-树5. File Transfer (25) 时间限制 150 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作者 CHEN, Yue We have ...
- PTA 04-树5 Root of AVL Tree (25分)
题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/668 5-6 Root of AVL Tree (25分) An AVL tree ...
- PTA 10-排序5 PAT Judge (25分)
题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/677 5-15 PAT Judge (25分) The ranklist of PA ...
- PTA 05-树7 堆中的路径 (25分)
题目地址 https://pta.patest.cn/pta/test/15/exam/4/question/713 5-5 堆中的路径 (25分) 将一系列给定数字插入一个初始为空的小顶堆H[] ...
- pta 编程题13 File Transfer
其它pta数据结构编程题请参见:pta 这道题考察的是union-find并查集. 开始把数组中每个元素初始化为-1,代表没有父节点.为了使树更加平衡,可以让每一个连通分量的树根的负值代表这个连通分量 ...
随机推荐
- bash 变量传递方法
###1.sh ##(该sh 目的是 将变量env传入env.sh, 同时让env.sh在当前事物生效,最后执行env.sh 定义的变量envs) export ENV=prepareecho ...
- RPC之远程过程调用
一. 简介 将一个函数运行在远程计算机上并且等待获取那里的结果,这个称作远程过程调用(Remote Procedure Call)或者 RPC. RPC是一个计算机通信协议. 1. 类比: 将计算机服 ...
- Android开发实现高德地图定位
1.获取Key 参考官方文档:http://lbs.amap.com/api/android-location-sdk/guide/create-project/get-key 对于签名文件的获取建议 ...
- java实现排序的几种方法
package com.ywx.count; import java.util.Scanner; /** * 题目:排序的几种方式(汇总及重构) * @author Vashon(yangwenxue ...
- RHEL5.8上SAMBA源码修改打包安装流程
之前一直使用系统自带的SAMBA,近期需要对SAMBA代码做一些修改,然后还是打算用RPM包的方式来安装部署. 这个流程本身不复杂,在这里记录下来,免得在另外写说明文档. 关键词:RHEL5.8, s ...
- phpstorm 格式化代码
MAC 安装phpcs.phpcbf composer global require 'squizlabs/php_codesniffer=*' Changed current directory t ...
- DBMS的工作模式
数据库管理系统(DBMS)是指数据库系统中对数据进行管理的软件系统,它是数据库系统的核心组成部分,对数据库的一切操作(增删改查)都是通过DBMS进行的 DBMS的工作模式如下: 1>接受应用程序 ...
- ChromiumWebBrowser禁止鼠标右键和拖动
在屏蔽之前先查看namespace CefSharp.WinForms内的代码 public class ChromiumWebBrowser : Control, IWebBrowserIntern ...
- Unity3D——Epirome框架_TimerManager计时任务管理器
1.Timer timer = new Timer(); 创建时间管理器 参数(float time, TimeUnit timeUnit,bool ignoreTimeScale = false, ...
- uva1380 A Scheduling Problem
按紫书来注意这道题的题目给了很大的方便,就相当于验证k是不是答案,不是的话就是k+1 #include<iostream> #include<string> #include& ...