HDU 5638 Toposort 拓扑排序 优先队列

Toposort

题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=5638

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

Hint

题意

给一个DAG，然后让你最多删除k条边，使得这个图的拓扑序最小。

题解：

贪心的想一想，现在我扔出来的点是一定是入度小于等于k，且编号最小的点。

这个怎么做呢？

线段树内二分，或者直接优先队列就好了。

choose what you like.

代码

#include<stdio.h>
#include<iostream>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int maxn = 2e5+7;
const int mod = 1e9+7;
vector<int> E[maxn],rE[maxn];
int in[maxn];
int inq[maxn];
int vis[maxn];
priority_queue<int,vector<int>,greater<int> >Q;
void init()
{
    for(int i=0;i<maxn;i++)
        E[i].clear(),rE[i].clear(),in[i]=0;
    memset(inq,0,sizeof(inq));
    memset(vis,0,sizeof(vis));
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        init();
        int n,m,k;
        scanf("%d%d%d",&n,&m,&k);
        for(int i=0;i<m;i++)
        {
            int x,y;scanf("%d%d",&x,&y);
            E[x].push_back(y);
            rE[y].push_back(x);
            in[y]++;
        }
        long long Ans = 0;
        for(int i = 1 ; i <= n ; ++ i)
        {
            if(in[i]<=k)
            {
                Q.push( i );
                inq[i] = 1;
            }
        }
        int num = 1;
        while(!Q.empty()){
            int x = Q.top() ; Q.pop(); inq[x] = 0;
            if(k >= in[x]){
                vis[x] = 1 , k -= in[x];
                Ans=(Ans+1ll*num*x)%mod;
                num=num+1;
                for(int i=0;i<E[x].size();i++){
                    int v =E[x][i];
                    if(vis[v]) continue;
                    in[v]--;
                    if(in[v] <= k&&!inq[v]){
                        Q.push(v);
                        inq[v] = 1;
                    }
                }
            }
        }
        printf("%I64d\n",Ans);
    }
}