HDU 6090-Rikka with Graph

Rikka with Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 821    Accepted Submission(s): 474

题目链接：点击打开链接

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

题意：

给你n个节点，让你选择m条边相连，dis[i][j]定义为i点和j点之间的最短路径，如果两个点被连接，则距离为1，如果有孤立的点，在 孤立的点到其他点的距离为n，其他点到孤立点的距离也为n，求图的最小权重，权重定义为：

分析：

本题为找规律的题。
(1).n个节点最多可以构成n*(n-1)条边，里面的每一条边都是双向的，所以当m>=n*(n-1)/2，ans=n*(n-1);
(2).当m>=(n-1),正好让n个节点构成一个连通图，此时以某一个点为中心点其他点都与之相连能够使得权重最小，

ans=n * (n - 1)+2 * (n * (n - 1) / 2 - m)，因为找出规律在不漏点的情况下，每少一条边数值就+2；
(3).当m<n-1,存在着游离的点，m条边，也就是说有m+1个点是联通的，n-m-1个点是游离的，ans=2*m*m+(m*n+n*n)*(n-1-m); 

对于第三种情况，在这里和小伙伴门解释一下，看图

#include <iostream>using namespace std;int main(){    int t;    cin >> t;    while (t--)    {        long long n, m;        cin >> n >> m;        if (n * (n - 1) / 2 <= m)        {            cout << n * (n - 1) << endl;            continue;        }        long long ans = n * (n - 1);        if (m >= (n - 1))        {            ans += 2 * (n * (n - 1) / 2 - m);        }        else        {            ans =2*m*m+(m*n+n*n)*(n-1-m);        }    cout << ans << endl;    }    return 0;}