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

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 mostk 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

5 5 21 24 52 43 42 33 2 01 21 3

 

Sample Output

5 3 1 2 41 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;}