hdu5154 Harry and Magical Computer--拓扑排序

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本文介绍了一道关于哈利波特魔法电脑的问题，通过图论的方法来判断一组任务是否存在依赖环，进而确定是否能顺利完成所有任务。文章提供了完整的AC代码实现。

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原题链接： http://acm.hdu.edu.cn/showproblem.php?pid=5154

一：原题内容

Problem 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≤n≤100,1≤m≤10000
The next following m lines, each line contains two numbers a b, indicates a dependencies (a, b).

1≤a,b≤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

Sample Output

  
   YES
NO

二：分析理解

判断是否有环。

三：AC代码

#include<iostream>    
#include<string.h>  
#include<algorithm>  

using namespace std;

struct Node
{
	int v;
	int next;
};

Node node[10005];
int inDegree[105];
int head[105];
int sta[105];
int num;
int n, m;
int a, b;

void Join(int b, int a)
{
	inDegree[a]++;
	node[num].v = a;
	node[num].next = head[b];
	head[b] = num++;
}

int main()
{
	while (~scanf("%d%d", &n, &m))//忘记了~，竟然卡了快1h，真是尿了，OJ真折磨人，这个折磨人的小妖精
	{
		memset(inDegree, 0, sizeof(inDegree));
		memset(head, -1, sizeof(head));
		num = 0;

		while (m--)
		{
			scanf("%d%d", &a, &b);
			Join(b, a);
		}

		num = 0;
		int top = 0;
		for (int i = 1; i <= n; i++)
		{
			if (inDegree[i] == 0)
			{
				num++;
				inDegree[i]--;
				sta[++top] = i;
			}
		}
		while (top)
		{
			int u = sta[top--];
			for (int i = head[u]; i != -1; i = node[i].next)
			{
				int v = node[i].v;
				inDegree[v]--;
				if (inDegree[v] == 0)
				{
					num++;
					sta[++top] = v;
					inDegree[v]--;
				}
			}
			
		}

		if (num == n)
			printf("YES\n");
		else
			printf("NO\n");
	}
	
	return 0;
}