hdu 5195 DZY Loves Topological Sorting（线段树）

DZY Loves Topological Sorting

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1065    Accepted Submission(s): 328

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).

题意：求删除k条边后字典序最大的拓扑排序

思路：一开始想在拓扑排序里面改，可是怎么也贪心不出来。

后面才知道能用线段树去维护，每次取入度<=k的边，并且编号尽量大（也就是尽量往右取），然后每次取完更新即可。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;
#define N 100005
#define INF 999999999
struct Edge
{
    int v,next;
}edge[N];
int tree[N<<2];
int num[N],cnt,head[N];
int ans,k;
void init()
{
    cnt=0;
    memset(num,0,sizeof(num));
    memset(head,-1,sizeof(head));
}
void addedge(int u,int v)
{
    edge[cnt].v=v;
    edge[cnt].next=head[u];
    head[u]=cnt++;
}
void pushup(int root)
{
    tree[root]=min(tree[root<<1],tree[root<<1|1]);
}
void build(int root,int l,int r)
{
    if(l==r)
    {
        tree[root]=num[l];
        return;
    }
    int mid=(l+r)>>1;
    build(root<<1,l,mid);
    build(root<<1|1,mid+1,r);
    pushup(root);
}
void query(int root,int l,int r)
{
    if(l==r)
    {
        k-=tree[root];
        ans=l;
        tree[root]=INF;
        return;
    }
    int mid=(l+r)>>1;
    if(tree[root<<1|1]<=k) query(root<<1|1,mid+1,r);
    else query(root<<1,l,mid);
    pushup(root);
}
void update(int pos,int root,int l,int r)
{
    if(l==r)
    {
        tree[root]--;
        return;
    }
    int mid=(l+r)>>1;
    if(mid>=pos) update(pos,root<<1,l,mid);
    else update(pos,root<<1|1,mid+1,r);
    pushup(root);
}
int main()
{
    int n,m;
    int u,v;
    while(~scanf("%d %d %d",&n,&m,&k))
    {
        init();
        for(int i=1; i<=m; i++)
                {
                    scanf("%d %d",&u,&v);
                    addedge(u,v);
                    num[v]++;
                }
        build(1,1,n);
        for(int i=1;i<=n;i++)
        {
            query(1,1,n);
            printf("%d",ans);
            if(i<n) printf(" ");
            else printf("\n");
            update(ans,1,1,n);
            for(int i=head[ans];i!=-1;i=edge[i].next)
            {
                int v=edge[i].v;
                num[v]--;
                update(v,1,1,n);
            }
        }
    }
    return 0;
}