hdu.5195.DZY Loves Topological Sorting(topo排序 && 贪心）

DZY Loves Topological Sorting

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 866    Accepted Submission(s): 250

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

5 5 2
1 2
4 5
2 4
3 4
2 3
3 2 0
1 2
1 3

 

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

 #include<stdio.h>
 #include<string.h>
 #include<queue>
 using namespace std;
 const int M =  ;
 struct Edge
 {
     int v , nxt ;
     Edge () {}
     Edge (int v , int nxt) : v (v) , nxt (nxt) {}
 }e[M];
 int H[M] , E ;
 bool vis[M] ;
 int ans[M] , top ;
 int in[M] ;
 int n , m , k ;
 int u , v ;
 inline int read () {
     int ans = ; char c; bool flag = false;
     while ((c = getchar()) == ' ' || c == '\n' || c == '\r');
     if (c == '-') flag = true; else ans = c - '';
     while ((c = getchar()) >= '' && c <= '') ans = ans *  + c - '';
     return ans * (flag ? - : );
 }

 void addedge ()
 {
     e[E] = Edge ( v , H[u] ) ;
     H[u] = E ++ ;
 }

 void init ()
 {
     E =  ;
     top =  ;
     memset (H , -  , sizeof(H)) ;
     memset (ans ,  , sizeof(ans) ) ;
     memset (in ,  , sizeof(in)) ;
     memset (vis ,  , sizeof(vis) ) ;
 }

 void topo ()
 {
     priority_queue <int> q ;
     while (!q.empty ()) q.pop () ;
     for (int i =  ; i <= n ; i++) if (in[i] ==  && !vis[i]) q.push (i) ;
     while ( !q.empty () ) {
         int u = q.top () ;
         q.pop () ;
         ans[top ++] = u ;
         for (int i = H[u] ; ~ i ; i = e[i].nxt) {
             in[e[i].v] -- ;
             if (in[e[i].v] ==  && !vis[e[i].v])    q.push (e[i].v) ;
         }
     }
 }

 void solve ()
 {
     init () ;
     while (m--) {
         u = read () , v = read () ;
         addedge () ;
         in[v] ++ ;
     }
     priority_queue <int> q ;
     for (int i =  ; i <= n ; i++)  if (in[i] <= k) q.push (i) ;
     while ( !q.empty () ) {
         u = q.top () ;
         q.pop () ;
         if (in[u] > k) continue ;
         ans[top ++] = u ;
         k -= in[u] ;
         vis[u] =  ;
         for (int i = H[u] ; ~ i ; i = e[i].nxt) {
            in[e[i].v] -- ;
            if (in[e[i].v] <= k && !vis[u] ) q.push (e[i].v) ;
         }
     }
     topo () ;
     for (int i =  ; i < top ; i++) {
         printf ("%d%c" , ans[i] , i == top -  ? '\n' : ' ') ;
     }
 }

 int main ()
 {
    // freopen ("a.txt" , "r" , stdin ) ;
     while (~ scanf ("%d%d%d" , &n , &m , &k)) {
         solve () ;
     }
     return  ;
 }

用邻接表优化后，topo的时间复杂度O(n),空间复杂度也大大减少。orz。
还有快速读入。

托它们的福，这道题让我530ms过了。233333333