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 = 100005 ;
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 = 0; char c; bool flag = false;
    while ((c = getchar()) == ' ' || c == '
' || c == '
');
    if (c == '-') flag = true; else ans = c - '0';
    while ((c = getchar()) >= '0' && c <= '9') ans = ans * 10 + c - '0';
    return ans * (flag ? -1 : 1);
}

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

void init ()
{
    E = 0 ;
    top = 0 ;
    memset (H , - 1 , sizeof(H)) ;
    memset (ans , 0 , sizeof(ans) ) ;
    memset (in , 0 , sizeof(in)) ;
    memset (vis , 0 , sizeof(vis) ) ;
}

void topo ()
{
    priority_queue <int> q ;
    while (!q.empty ()) q.pop () ;
    for (int i = 1 ; i <= n ; i++) if (in[i] == 0 && !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] == 0 && !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 = 1 ; 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] = 1 ;
        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 = 0 ; i < top ; i++) {
        printf ("%d%c" , ans[i] , i == top - 1 ? '
' : ' ') ;
    }
}

int main ()
{
   // freopen ("a.txt" , "r" , stdin ) ;
    while (~ scanf ("%d%d%d" , &n , &m , &k)) {
        solve () ;
    }
    return 0 ;
}

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