Legal or Not（拓扑排序）

最新推荐文章于 2024-08-29 11:53:56 发布

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最新推荐文章于 2024-08-29 11:53:56 发布

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分类专栏： C算法

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本文介绍了一种利用拓扑排序算法来判断输入数据是否形成非法闭环的方法。通过构建图结构并进行拓扑排序，可以有效地检测出数据之间的依赖关系是否合法。

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ACM-DIY is a large QQ group where many excellent acmers get together. It is so harmonious that just like a big family. Every day,many “holy cows” like HH, hh, AC, ZT, lcc, BF, Qinz and so on chat on-line to exchange their ideas. When someone has questions, many warm-hearted cows like Lost will come to help. Then the one being helped will call Lost “master”, and Lost will have a nice “prentice”. By and by, there are many pairs of “master and prentice”. But then problem occurs: there are too many masters and too many prentices, how can we know whether it is legal or not?

We all know a master can have many prentices and a prentice may have a lot of masters too, it’s legal. Nevertheless，some cows are not so honest, they hold illegal relationship. Take HH and 3xian for instant, HH is 3xian’s master and, at the same time, 3xian is HH’s master,which is quite illegal! To avoid this,please help us to judge whether their relationship is legal or not.

Please note that the “master and prentice” relation is transitive. It means that if A is B’s master ans B is C’s master, then A is C’s master.

Input
The input consists of several test cases. For each case, the first line contains two integers, N (members to be tested) and M (relationships to be tested)(2 <= N, M <= 100). Then M lines follow, each contains a pair of (x, y) which means x is y’s master and y is x’s prentice. The input is terminated by N = 0.

TO MAKE IT SIMPLE, we give every one a number (0, 1, 2,…, N-1). We use their numbers instead of their names.

Output
For each test case, print in one line the judgement of the messy relationship.

If it is legal, output “YES”, otherwise “NO”.

Sample Input
3 2
0 1
1 2
2 2
0 1
1 0
0 0
Sample Output
YES
NO

这个题就是通过拓扑排序判断输入数据是否组成环。
如果成环，所给的数据就不能全部输出，我们可以设置一个变量来判断。

#include<cstdio>
#include<vector>
#include<queue>
#include<cstring>
#define MAX 205
using namespace std;
vector<int>vec[MAX];//储存输入元素
queue<int>q;//把判断出度为0的点存入队列
int zero[MAX];//储存度的值
int n,m;
int num;//用这个判断储存的点的个数
bool top_sort()
{
while(!q.empty())//每调用一次函数要先清空队列
q.pop();
num=0;
for(int i=0;i<n;i++)
if(!zero[i])//如果度为0，进入队列
q.push(i);
while(!q.empty())//队列不为空的时候
{
	int now=q.front();//找队列中第一个元素
	q.pop();//首元素出队
	num++;//统计度为0的元素
	for(int i=0;i<vec[now].size();i++)
	{
		if(--zero[vec[now][i]]==0)//继续寻找度为0的点
		q.push(vec[now][i]);
	}
}
if(num==n)
return true;
else
return false;
}
int main()
{
int x,y;
while(scanf("%d %d",&n,&m)!=EOF&&(m||n))
{
	for(int i=0;i<n;i++)
	vec[i].clear();//清空数组
	memset(zero,0,sizeof(zero));
	for(int i=0;i<m;i++)
	{
		scanf("%d %d",&x,&y);//输入数据
		vec[x].push_back(y);//建立方向关系
		zero[y]++;//度增加
	}
	if(top_sort())
	{
		printf("YES\n");
	}
	else
	printf("NO\n");
}
return 0;
}