HDU 5638 Toposort

Problem Description

There is a directed acyclic graph with n vertices and m edges. You are allowed to delete exact k edges in such way that the lexicographically minimal topological sort of the graph is minimum possible.

 

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains three integers n, m and k (1≤n≤100000,0≤k≤m≤200000) -- the number of vertices, the number of edges and the number of edges to delete.

For the next m lines, each line contains two integers ui and vi, which means there is a directed edge from ui to vi (1≤ui,vi≤n).

You can assume the graph is always a dag. The sum of values of n in all test cases doesn't exceed 106. The sum of values of m in all test cases doesn't exceed 2×106.

 

Output

For each test case, output an integer S=(∑i=1ni⋅pi) mod (109+7), where p1,p2,...,pn is the lexicographically minimal topological sort of the graph.

 

Sample Input

3
4 2 0
1 2
1 3
4 5 1
2 1
3 1
4 1
2 3
2 4
4 4 2
1 2
2 3
3 4
1 4

 

Sample Output

30
27
30

 

有向图删除k条边使得拓扑序列字典序最小，可以直接用优先队列维护。

#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<functional>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn=1e5+10;
const int mod=1e9+7;
int T,n,m,k,cnt[maxn],x,y,vis[maxn];
vector<int> t[maxn];

int main()
{
	scanf("%d",&T);
	while (T--)
	{
		scanf("%d%d%d",&n,&m,&k);
		for (int i=1;i<=n;i++) t[i].clear(),cnt[i]=vis[i]=0;
		while (m--)
		{
			scanf("%d%d",&x,&y);
			t[x].push_back(y);
			cnt[y]++;
		}
		priority_queue<int,vector<int>,greater<int> > p;
		for (int i=1;i<=n;i++) if (cnt[i]<=k) p.push(i),vis[i]=1;
		int res=0,i=1;
		while (!p.empty())
		{
			int q=p.top();	p.pop();
			if (cnt[q]>k) {vis[q]=0; continue;}
			(res+=(LL)q*i++%mod)%=mod;
			k-=cnt[q];
			for (int i=0;i<t[q].size();i++)
			{
				cnt[t[q][i]]--;
				if (!vis[t[q][i]]&&cnt[t[q][i]]<=k)
				{
					p.push(t[q][i]);
					vis[t[q][i]]=1;
				}
			}
		}
		printf("%d\n",res);
	}
	return 0;
}