codeforces Kitchen Plates （拓扑或者暴力）

最新推荐文章于 2021-12-27 15:46:25 发布

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本文探讨了Codeforces上的一道题目，涉及到通过一系列比较关系来确定五个不同大小厨房盘子的正确排序。文章提供了使用拓扑排序解决此问题的代码实现，同时对比了暴力解法next_permutation()的效率。

http://codeforces.com/gym/102219/problem/J

Kitchen Plates

time limit per test

1.0 s

memory limit per test

256 MB

input

standard input

output

standard output

You are given 5 different sizes of kitchen plates. Each plate is marked with a letter AA, BB, CC, DD, or EE. You are given 5 statements comparing two different plates, you need to rearrange the plates from smallest size to biggest size. For example: the sizes of these plates.

Input

The input consist of 5 lines. In each line there will be 3 characters, the first and last character will be either AA, BB, CC, DD, or EE and the middle character will be either >> or << describing the comparison between two plates sizes. No two plates will be equal.

Output

The output consist of 55 characters, the sorted order of balls from smallest to biggest plate. Otherwise, if the statements are contradicting print impossibleimpossible. If there are multiple answers, print any of them.

Examples

input

Copy

D>B
A>D
E<C
A>B
B>C

output

Copy

ECBDA

input

Copy

B>E
A>B
E>A
C<B
D<B

output

Copy

impossible

这题写了半天拓扑，别人用next_permutation( ) 暴力十分钟写出来了

心拔凉拔凉的

关于拓扑排序：https://blog.youkuaiyun.com/qq_41431457/article/details/89787036

#include<bits/stdc++.h>
using namespace std;

map<char,int> M;
bool vis[6], dir[6][6];
int in[6];
vector<int>  V;
char f[5] = {'A', 'B','C','D','E'};
priority_queue<int,vector<int>,greater<int> > Q;
void toposort()
{
	for(int i=0;i<5;i++)//将所有入度为0的顶点放入序列 
	if(in[i] == 0 ) Q.push(i);
//	cout << Q.size() <<endl;
	while(!Q.empty())
	{
		int x=Q.top(); Q.pop();
		V.push_back(x);
		
		for(int i=0;i<5;i++)//删除所有与x顶点有关的边 
		if(dir[x][i])
		{
			in[i]--;//邻接点入度减一 
			if(!in[i])
			Q.push(i);
		}
	}
}

int main()
{
	M['A'] = 0;
	M['B'] = 1;
	M['C'] = 2;
	M['D'] = 3;
	M['E'] = 4;
	
	int k = 5;
	while(k -- )
	{
		char a,b,c;
		cin >> a >> b >> c;
		if(b == '<') swap(a,c);
		
		in[M[c]]++;
		
		dir[M[a]][M[c]] = vis[M[a]] = vis[M[c]] = true;
	}
	
	toposort();
	
	if(V.size() < 5) 
	{
		cout << "impossible" << endl;
		return 0;
	}
	
	for(int i = 4; i >= 0; i--)
	cout << f[V[i]];
	cout<<endl;
	
	return 0;
}