Bestcoder Round #35 (dzy Loves topological sorting-heap + greedy)

Dzy Loves Topological sortingTime limit:4000/2000 MS (java/others) Memory limit:131072/131072 K (java/others)
Total submission (s): 323 Accepted submission (s): 86


Problem Descriptiona topological sort or topological ordering of a directed graph is a linear ordering of its vertices suc h that for every directed edge ( Span id= "mathjax-span-4" class= "Mi" style= "font-family:mathjax_math-italic-web" >u → v ) From vertex u to vertex v , u Comes before v In the ordering.
Now, the Dzy has a directed acyclic graph (DAG). Should find the lexicographically largest topological ordering after erasing at most k Edges from the graph.
Inputthe input consists several test cases. ( T esTC ase≤5 )
The first line, three integers N,m,k(1≤N,m≤ ten 5 ,0≤k≤m) .
Each of the next m Lines has integers: u,v(u≠v,1≤u,v≤N) , representing a direct edge (u→v) .
Outputfor 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 3 2Hintcase 1.Erase The Edge (2->3), (4->5). and the lexicographically largest topological ordering is (5,3,1,2,4).

Sourcebestcoder Round #35
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Direct heap + greedy.

The next choice is I, obviously.

When and only if I is indegree<= the remaining number of deleted Nowk, the number of the largest

So build a heap, throw the indegree<=k in, roll out the maximum number,

Worth a question is I did not calculate the complexity, and finally directly past.

Maybe Nowk down, is going to put the original heap in the element t out, but I failed to find the inverse of tle

PS: Because next is an existing function in the system, it cannot be used,

#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)/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];p Re[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;}

Bestcoder Round #35 (dzy Loves topological sorting-heap + greedy)