BestCoder Round #35(DZY Loves Topological Sorting-堆+贪心)

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DZY Loves Topological Sorting

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 323 Accepted Submission(s): 86

Problem Description

A topological sort or topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge

(u→v) from vertex

u to vertex

v ,

u comes before

v in the ordering.
Now, DZY has a directed acyclic graph(DAG). You should find the lexicographically largest topological ordering after erasing at most

k edges from the graph.

Input

The input consists several test cases. (

TestCase≤5 )
The first line, three integers

n,m,k(1≤n,m≤105,0≤k≤m) .
Each of the next

m lines has two integers:

u,v(u≠v,1≤u,v≤n) , representing a direct edge

(u→v) .

Output

For each test case, output the lexicographically largest topological ordering.

Sample Input

Sample Output

  
   5 3 1 2 4
1 3 2


   
    
     Hint
    Case 1.
Erase the edge (2->3),(4->5).
And the lexicographically largest topological ordering is (5,3,1,2,4).

Source

BestCoder Round #35

Recommend

hujie | We have carefully selected several similar problems for you: 5197 5196 5193 5192 5191

直接堆+贪心。

显然下一个选取的是i，

当且仅当i是indegree<=剩余删边数nowk，中编号最大的

因此建一个堆，把indegree<=k的扔进去，推出最大编号，

值得一题的是我没算过复杂度，最后直接过了。。

可能nowk降低，是会把原来堆中的元素T出来，但我未能找出TLE的反例

PS：因为next是系统中已有函数所以不能用，，

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
#include<queue>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (200000+10)
#define MAXM (200000+10) 
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
priority_queue<int> q;
bool b[MAXN]={0};
int indegree[MAXN]={0};

int edge[MAXM],pre[MAXN],Next[MAXM],siz=0;
void addedge(int u,int v)
{
	edge[++siz]=v;
	Next[siz]=pre[u];
	pre[u]=siz; 
}

int n,m,k;

int main()
{
//	freopen("Sorting.in","r",stdin);
	
	while(scanf("%d%d%d",&n,&m,&k)==3)
	{
		while(!q.empty()) q.pop();
		MEM(b) MEM(indegree) MEM(edge) MEM(pre) MEM(Next) siz=0;	 
		
		
		For(i,m)
		{
			int u,v;
			scanf("%d%d",&u,&v);
			addedge(u,v);
			indegree[v]++;
		}
		
		For(i,n)
		{
			if (indegree[i]<=k) b[i]=1,q.push(i);
		}
		
		int nowk=k;
		int flag=0;		
		while(!q.empty())
		{
			int t;
			while(1)
			{
				t=q.top();q.pop();
				if (indegree[t]>nowk) b[t]=0;
				else break;
			}
			
			nowk-=indegree[t];indegree[t]=0;
			
			Forp(t)
			{
				int v=edge[p];
				indegree[v]--;
				if (indegree[v]<=nowk&&b[v]==0) b[v]=1,q.push(v); 
			}							
			
			if (flag) printf(" ");flag++; 
			printf("%d",t);
			if (flag==n) break;
			
		} 
		putchar('\n');
			
		
	}	
	
	return 0;
}