2017杭电多校06Rikka with Graph

Rikka with Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 2377    Accepted Submission(s): 766

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

14

 

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题意：给出n个顶点，m条边，要求图的最小距离

思路：题中给出两点距离的值是经过的边数，所以尽可能的把所有的点连成菊花链；

分三种情况讨论：

1.当m<=n-1

分别计算连通点，孤立点，连通与孤立三种情况

result=(m+1-1)*(m+1-1)*2+(n-m-1)*(n-m-1)*n+(m+1)*(n-m-1)*2*n;

2.当m>n-1&&m<n*(n-1)/2

转换为边数为n*(n-1)/2的图的基础上删掉了n*(n-1)/2-m条边，每删掉一条边，距离数加上2

result=n*(n-1)+(n*(n-1)/2-m)*2；

3.当m>=n*(n-1)/2

显然此类情况图中所有的点都能直接相连，所以result=n*(n-1)；

代码如下：

#include <iostream>

using namespace std;
typedef long long ll;
int main()
{
    int t;
    scanf("%d",&t);
    ll n,m;
    while(t--)
    {
        scanf("%lld%lld",&n,&m);
        ll tmp=n*(n-1)/2;
        ll result=0;
        if(m>=tmp)
        {
            result=n*(n-1);
        }
        else if(m>n-1&&m<tmp)
        {
            result=(n*(n-1))+(tmp-m)*2;
        }
        else
        {
            ll p=m+1,q=n-m-1;
            result=m*m*2+(n-m-1)*(n-m-2)*n+p*q*2*n;
        }
        cout<<result<<endl;
    }
    return 0;
}