Rikka with Graph（HDU 6090）

Rikka with Graph

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

Total Submission(s): 703    Accepted Submission(s): 411

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：已知一张图里有n个顶点，m条边，求每个点到其他各点的距离之和，再求这些和的和的最小值。每条边权值为1，不连通的2点间距离为n。

//思路：分3种情况讨论：

(1) m>= n*(n-1)/2

这种情况下，每个点必然与其他所有点之间都有至少1条边，所以sum=n*(n-1)*1;

(2)n-1<=m<n*(n-1)/2

这种情况下，每两个点之间不一定有边，但我们通过找规律可得在完全图的基础上没少1条边，sum至少会增加2，所以可得：sum=n*(n-1)+(n*(n-1)/2 - m)*2;

(3)0<m<n-1

这种情况下必然会有孤立点，我们把尽可能多的点连在一起，因为孤立是最差的情况。然后就得到一个满足(2)的顶点数为m+1的图，和(n-m-1)个孤立点。sum=n*(n-1)*(n-m-1)+n*(n-m-1)*(m+1)+m*(m+1)+(m*(m+1)/2 - m)*2;

(孤立点到其他所有点的距离为n，其他所有点到孤立点的距离也为n)

还有数据要long long，int会爆，n虽然是1e6，但下面要算n^2，是1e12，用int会WA。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <algorithm>
using namespace std;

long long n;
long long m;

int main()
{
	int T;
	scanf("%d", &T);
	while (T--)
	{
		scanf("%lld%lld", &n, &m);
		long long sum = 0;
		if (m >= (n*(n - 1) / 2))
		{
			sum = n*(n - 1);
		}
		else if (m >= (n - 1) && m < (n*(n - 1) / 2))
		{
			sum = n*(n - 1) + (n*(n - 1) / 2 - m) * 2;
		}
		else if (m > 0 && m < (n - 1))
		{
			sum = n*(n - 1)*(n - m - 1) + n*(m + 1)*(n - m - 1) + m*(m + 1) + (m*(m + 1) / 2 - m) * 2;
		}
		printf("%lld\n", sum);
	}
	return 0;
}