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
1 #include<stdio.h> 2 #include<string.h> 3 #include<queue> 4 using namespace std; 5 const int M = 100005 ; 6 struct Edge 7 { 8 int v , nxt ; 9 Edge () {} 10 Edge (int v , int nxt) : v (v) , nxt (nxt) {} 11 }e[M]; 12 int H[M] , E ; 13 bool vis[M] ; 14 int ans[M] , top ; 15 int in[M] ; 16 int n , m , k ; 17 int u , v ; 18 inline int read () { 19 int ans = 0; char c; bool flag = false; 20 while ((c = getchar()) == ' ' || c == '\n' || c == '\r'); 21 if (c == '-') flag = true; else ans = c - '0'; 22 while ((c = getchar()) >= '0' && c <= '9') ans = ans * 10 + c - '0'; 23 return ans * (flag ? -1 : 1); 24 } 25 26 void addedge () 27 { 28 e[E] = Edge ( v , H[u] ) ; 29 H[u] = E ++ ; 30 } 31 32 void init () 33 { 34 E = 0 ; 35 top = 0 ; 36 memset (H , - 1 , sizeof(H)) ; 37 memset (ans , 0 , sizeof(ans) ) ; 38 memset (in , 0 , sizeof(in)) ; 39 memset (vis , 0 , sizeof(vis) ) ; 40 } 41 42 void topo () 43 { 44 priority_queue <int> q ; 45 while (!q.empty ()) q.pop () ; 46 for (int i = 1 ; i <= n ; i++) if (in[i] == 0 && !vis[i]) q.push (i) ; 47 while ( !q.empty () ) { 48 int u = q.top () ; 49 q.pop () ; 50 ans[top ++] = u ; 51 for (int i = H[u] ; ~ i ; i = e[i].nxt) { 52 in[e[i].v] -- ; 53 if (in[e[i].v] == 0 && !vis[e[i].v]) q.push (e[i].v) ; 54 } 55 } 56 } 57 58 void solve () 59 { 60 init () ; 61 while (m--) { 62 u = read () , v = read () ; 63 addedge () ; 64 in[v] ++ ; 65 } 66 priority_queue <int> q ; 67 for (int i = 1 ; i <= n ; i++) if (in[i] <= k) q.push (i) ; 68 while ( !q.empty () ) { 69 u = q.top () ; 70 q.pop () ; 71 if (in[u] > k) continue ; 72 ans[top ++] = u ; 73 k -= in[u] ; 74 vis[u] = 1 ; 75 for (int i = H[u] ; ~ i ; i = e[i].nxt) { 76 in[e[i].v] -- ; 77 if (in[e[i].v] <= k && !vis[u] ) q.push (e[i].v) ; 78 } 79 } 80 topo () ; 81 for (int i = 0 ; i < top ; i++) { 82 printf ("%d%c" , ans[i] , i == top - 1 ? '\n' : ' ') ; 83 } 84 } 85 86 int main () 87 { 88 // freopen ("a.txt" , "r" , stdin ) ; 89 while (~ scanf ("%d%d%d" , &n , &m , &k)) { 90 solve () ; 91 } 92 return 0 ; 93 }
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
