【拓扑排序】【线段树】Gym

Consider a directed graph G of N nodes and all edges (u→v) such that u < v. It is clear that this graph doesn’t contain any cycles.

Your task is to find the lexicographically largest topological sort of the graph after removing a given list of edges.

A topological sort of a directed graph is a sequence that contains all nodes from 1 to N in some order such that each node appears in the sequence before all nodes reachable from it.

Input

The first line of input contains a single integer T, the number of test cases.

The first line of each test case contains two integers N and M (1 ≤ N ≤ 10⁵) , the number of nodes and the number of edges to be removed, respectively.

Each of the next M lines contains two integers a and b (1 ≤ a < b ≤ N), and represents an edge that should be removed from the graph.

No edge will appear in the list more than once.

Output

For each test case, print N space-separated integers that represent the lexicographically largest topological sort of the graph after removing the given list of edges.

Example

Input

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

Output

3 1 2
1 2 3 4
2 3 1 4

把普通的拓扑排序的栈操作改成线段树区间减一，查询区间最右侧的0的位置即可。注意一开始就删除的边，在区间减后要单点加回来。
然后当前的点处理完后，要把其置成无穷大。具体看代码。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 100010
int T,n,m,minv[N<<2],delta[N<<2];
void pushdown(int rt)//将rt结点的懒惰标记下传 
{
	if(delta[rt])
	  {
		delta[rt<<1]+=delta[rt];//标记下传到左结点 
		delta[rt<<1|1]+=delta[rt];//标记下传到右结点 
		minv[rt<<1]+=delta[rt];
		minv[rt<<1|1]+=delta[rt];
		delta[rt]=0;//清除rt结点的标记，下传完成 
	  }
}
void update(int ql,int qr,int v,int rt,int l,int r)
{
	if(ql<=l&&r<=qr)
	  {
		delta[rt]+=v;//更新当前结点的标记值 
		minv[rt]+=v;
		return ;
	  }
	pushdown(rt);//将该节点的标记下传到孩子们 
	int m=(l+r>>1);
	if(ql<=m) update(ql,qr,v,rt<<1,l,m);
	if(m<qr) update(ql,qr,v,rt<<1|1,m+1,r);
	minv[rt]=min(minv[rt<<1],minv[rt<<1|1]);
}
int query(int rt,int l,int r)
{
	if(l==r)
	  return l;
	int m=(l+r>>1);
	pushdown(rt);
	if(minv[rt<<1|1]==0) return query(rt<<1|1,m+1,r);
	else return query(rt<<1,l,m);
}
int first[N],e,next[N],v[N];
void AddEdge(int U,int V)
{
	v[++e]=V;
	next[e]=first[U];
	first[U]=e;
}
int del[N];
int main()
{
	int x,y;
	scanf("%d",&T);
	for(;T;--T)
	  {
	  	e=0;
	  	memset(minv,0,sizeof(minv));
	  	memset(delta,0,sizeof(delta));
	  	memset(first,0,sizeof(first));
	  	memset(next,0,sizeof(next));
	  	memset(del,0,sizeof(del));
	  	memset(v,0,sizeof(v));
	  	scanf("%d%d",&n,&m);
	  	for(int i=1;i<=m;++i)
	  	  {
	  	  	scanf("%d%d",&x,&y);
	  	  	AddEdge(x,y);
	  	  	++del[y];
	  	  }
	  	for(int i=1;i<=n;++i)
	  	  update(i,i,i-1-del[i],1,1,n);
	  	for(int i=1;i<=n;++i)
	  	  {
	  	  	int U=query(1,1,n);
	  	  	printf("%d%c",U,i==n ? '
' : ' ');
	  	  	update(U,U,n+1,1,1,n);
	  	  	update(U+1,n,-1,1,1,n);
	  	  	for(int j=first[U];j;j=next[j])
	  	  	  update(v[j],v[j],1,1,1,n);
	  	  }
	  }
	return 0;
}