hdu 5154 Harry and Magical Computer
题目连接
http://acm.hdu.edu.cn/showproblem.php?pid=5154
Harry and Magical Computer
Description
In reward of being yearly outstanding magic student, Harry gets a magical computer. When the computer begins to deal with a process, it will work until the ending of the processes. One day the computer got n processes to deal with. We number the processes from 1 to n. However there are some dependencies between some processes. When there exists a dependencies (a, b), it means process b must be finished before process a. By knowing all the m dependencies, Harry wants to know if the computer can finish all the n processes.
Input
There are several test cases, you should process to the end of file.
For each test case, there are two numbers n m on the first line, indicates the number processes and the number of dependencies. $1 \leq n \leq 100,1 \leq m \leq 10000$
The next following m lines, each line contains two numbers a b, indicates a dependencies $(a, b). 1 \leq a, b \leq n$
Output
Output one line for each test case.
If the computer can finish all the process print "YES" (Without quotes).
Else print "NO" (Without quotes).
Sample Input
3 2
3 1
2 1
3 3
3 2
2 1
1 3
Sample Output
YES
NO
先建立一张有向图,再遍历,若有顶点未访问到输出"NO",否则输出"YES"。。。
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<map>
using std::cin;
using std::cout;
using std::endl;
using std::find;
using std::sort;
using std::pair;
using std::vector;
using std::queue;
using std::multimap;
#define pb(e) push_back(e)
#define sz(c) (int)(c).size()
#define mp(a, b) make_pair(a, b)
#define all(c) (c).begin(), (c).end()
#define iter(c) decltype((c).begin())
#define cls(arr,val) memset(arr,val,sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i)
const int N = ;
int n, m, G[N][N];
bool vis[N], flag[N];
typedef unsigned long long ull;
void bfs() {
bool f = false;
queue<int> que;
rep(i, n) {
if (!flag[i]) {
vis[i] = true;
que.push(i);
}
}
while (!que.empty()) {
int p = que.front(); que.pop();
rep(i, n) {
if (G[p][i]) {
if (vis[i]) continue;
que.push(i);
vis[i] = true;
}
}
}
rep(i, n) {
if (!vis[i]) { f = true; break; }
}
puts(f ? "NO" : "YES");
}
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w+", stdout);
#endif
int a, b;
while (~scanf("%d %d", &n, &m)) {
cls(vis, ), cls(flag, ), cls(G, );
rep(i, m) {
scanf("%d %d", &a, &b);
G[b - ][a - ] = ;
flag[a - ] = ;
}
bfs();
}
return ;
}
hdu 5154 Harry and Magical Computer的更多相关文章
- hdu 5154 Harry and Magical Computer 拓扑排序
Harry and Magical Computer Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Ja ...
- (简单) HDU 5154 Harry and Magical Computer,图论。
Description In reward of being yearly outstanding magic student, Harry gets a magical computer. When ...
- HDU 5154 Harry and Magical Computer bfs
Harry and Magical Computer Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Ja ...
- HDU 5154 Harry and Magical Computer 有向图判环
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5154 题解: 有向图判环. 1.用dfs,正在访问的节点标记为-1,已经访问过的节点标记为1,没有访 ...
- 【HDOJ】5154 Harry and Magical Computer
拓扑排序. /* 5154 */ #include <iostream> #include <cstdio> #include <cstring> #include ...
- BC Harry and Magical Computer (拓扑排序)
Harry and Magical Computer Accepts: 350 Submissions: 1348 Time Limit: 2000/1000 MS (Java/Others) ...
- hdu 5154 拓扑排序
例题:hdu 5154 链接 http://acm.hdu.edu.cn/showproblem.php?pid=5154 题目意思是第一行先给出n和m表示有n件事,m个关系,接下来输入m行,每行有 ...
- BestCoder25 1001.Harry and Magical Computer(hdu 5154) 解题报告
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5154 题目意思:有 n 门 processes(编号依次为1,2,...,n),然后给出 m 种关系: ...
- HDU5154拓扑排序
Harry and Magical Computer Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Ja ...
随机推荐
- 下载编译和测试Android 源代码
http://source.android.com/source/downloading.html 其中出现错误 repo: fatal: error unknown url type: https ...
- 【练习】sqlnet.ora
在SQLNET.ora文件中设置以下参数可以实现IP访问限制: $ pwd/u01/app/oracle/product/10.2.0/db_1/network/admin$ vi sqlnet.or ...
- Google Protocol Buffer 协议
1. Protocol Buffers 简介 Protocol Buffers (ProtocolBuffer/ protobuf )是Google公司开发的一种数据描述语言,类似于XML能够将结构化 ...
- CENTOS7设置显示中文
vi /etc/locale.conf LANG="zh_CN.UTF-8" LANGUAGE="zh_CN.GB18030:zh_CN.GB2312:zh_CN&quo ...
- Eclipse 启动Tomcat后web项目的classes的子文件夹中没有calss文件
Eclipse 启动Tomcat后web项目的classes的子文件夹中没有calss文件. 经网上查得以下方法可解决 把properties属性里的java compiler-->buildi ...
- WP8_ListBox的用法
在Windows Phone 7 Tips (5) 中曾经提到,在Windows Phone 7 中页面的布局一般分为:Panoramic.Pivot.List和Full Screen.而通常List ...
- WP8__从windowsphone app store 中根据app id获取应用的相关信息(下载网址及图片id等)
windows phone 官网应用商店地址 http://www.windowsphone.com/zh-cn/store/featured-apps------------------------ ...
- 读取Jar包中的资源问题探究
最近在写一个可执行jar的程序,程序中包含了2个资源包,一个是images,一个是files.问题来了,在Eclipse里开发的时候,当用File类来获取files下面的文件时,没有任何问题.但是当程 ...
- Backup App's data without rooting the phone
First I'd like to let you know that my phone is Android 6.0 Marshmallow. So it works on the latest A ...
- leetcode 121
121. Best Time to Buy and Sell Stock Say you have an array for which the ith element is the price of ...