hdu 5195 DZY Loves Topological Sorting【拓扑排序+优先队列+邻接表】

DZY Loves Topological Sorting

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1014    Accepted Submission(s): 303

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).

题目大意：

一张有向图的拓扑序列是图中点的一个排列，满足对于图中的每条有向边$(u\to v)$ 从 $u$ 到 $v$，都满足$u$在排列中出现在$v$之前。
现在，DZY有一张有向无环图(DAG)。你要在最多删去$k$条边之后，求出字典序最大的拓扑序列。

题目保证了图是合法的拓扑图、这里我们就不用多担心了。这里说要删除k条边，然后求出字典序最大的拓扑排序，看到这里，我们马上就应该有一种贪心的思想：去边的时候，找度小于等于k的点，使这个点的编号尽量大，然后去度、这里求的是字典序最大的拓扑序列，我们可以应用优先队列来搞定

因为图比较大，所以这里用vector来搞定图的问题：

#include<stdio.h>#include<string.h>#include<queue>#include<algorithm>#include<vector>using namespace std;int now,head[200010],nex[200010],p[200010];int vis[200010];int degree[200010];vector<int>map[200010];int ans[200010];/*void add(int x,int y){    nex[++now]=head[x];    head[x]=now;    p[now]=y;}*/int main(){    int n,m,k;    while(~scanf("%d%d%d",&n,&m,&k))    {        for(int i=1;i<=n;i++)        {            map[i].clear();        }        memset(vis,0,sizeof(vis));        memset(degree,0,sizeof(degree));        memset(head,0,sizeof(head));        now=0;        while(m--)        {            int x,y;            scanf("%d%d",&x,&y);            map[x].push_back(y);            //add(x,y);            degree[y]++;        }        priority_queue<int>s;        for(int i=1;i<=n;i++)        {            if(degree[i]<=k)            s.push(i);        }        int cont=0;        while(!s.empty())        {            int u=s.top();            s.pop();            if(vis[u]||degree[u]>k)continue;            vis[u]=1;            k-=degree[u];            ans[++cont]=u;            for(int j=0;j<map[u].size();j++)            {                int v=map[u][j];                degree[v]--;                if(degree[v]<=k&&vis[v]==0)                s.push(v);            }            /*            for(int i=head[u];i;i=nex[i])            {                int v=p[i];                degree[v]--;                if(degree[v]<=k&&vis[v]==0)                s.push(v);            }*/        }        for(int i=1;i<cont;i++)        {            printf("%d ",ans[i]);        }        printf("%d\n",ans[cont]);    }}