HDU 6090 17多校5 Rikka with Graph（思维简单题）

Posted 2020-10-01 Annetree的博客

Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

For an undirected graph G with n nodes and m edges, we can define the distance between (i,j) (dist(i,j)) as the length of the shortest path between i and j. The length of a path is equal to the number of the edges on it. Specially, if there are no path between i and j, we make dist(i,j) equal to n.

Then, we can define the weight of the graph G (wG) as ∑ni=1∑nj=1dist(i,j).

Now, Yuta has n nodes, and he wants to choose no more than m pairs of nodes (i,j)(i≠j) and then link edges between each pair. In this way, he can get an undirected graph G with n nodes and no more than m edges.

Yuta wants to know the minimal value of wG.

It is too difficult for Rikka. Can you help her?  

In the sample, Yuta can choose (1,2),(1,4),(2,4),(2,3),(3,4).

 

 

Input

The first line contains a number t(1≤t≤10), the number of the testcases. 

For each testcase, the first line contains two numbers n,m(1≤n≤106,1≤m≤1012).

 

 

Output

For each testcase, print a single line with a single number -- the answer.

 

 

Sample Input

1

4 5

 

 

Sample Output

1

4

 

分成三种情况考虑

1.m大大的有，超过了n*(n-1)/2的情况，n*(n-1)/2说明每个点之间都有连线，那就是最少的情况n*(n-1)

2.m可以把所有点连在一起。这样的话随便推算几个就可以发现规律了。

3.m不够，然后就把连在一起和不连在一起的分开算，具体看代码注释。

 

 1 #include <iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<queue>
 6 #include<cmath>
 7 #include<string>
 8 #include<map>
 9 #include<vector>
10 using namespace std;
11 
12 long long calc(long long n,long long m)
13 {
14     return 2*(n-1)*(n-1)-(m-n+1)*2;
15 }
16 
17 int main()
18 {
19     long long T,n,m,ans;
20     scanf("%d",&T);
21     while(T--)
22     {
23         scanf("%lld%lld",&n,&m);
24         if(m>=n*(n-1)/2)
25             ans=n*(n-1);
26         else if(m<n-1)
27             ans=calc(1+m,m)+(n-1-m)*(m+1)*2*n/*和连通的内部点连*/+(n-1-m)*(n-2-m)*n/*孤立的点之间*/;
28         else
29             ans=calc(n,m);
30         printf("%lld\n",ans);
31     }
32     return 0;
33 }