HDU

Rikka with Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 121    Accepted Submission(s): 86

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

14 5

 

Sample Output

14

 

Source

2017 Multi-University Training Contest - Team 5 

 

官方题解：

考虑贪心地一条一条边添加进去。

当 $m\le n-1$ 时，我们需要最小化距离为 $n$ 的点对数，所以肯定是连出一个大小为 $m+1$ 的联通块，剩下的点都是孤立点。在这个联通块中，为了最小化内部的距离和，肯定是连成一个菊花的形状，即一个点和剩下所有点直接相邻。

当 $m>n-1$ 时，肯定先用最开始 $n-1$ 条边连成一个菊花，这时任意两点之间距离的最大值是 $2$。因此剩下的每一条边唯一的作用就是将一对点的距离缩减为 $1$。

这样我们就能知道了最终图的形状了，稍加计算就能得到答案。要注意 $m$ 有可能大于 n(n−1)2​2​​n(n−1)​​。

#include<iostream>#include<string.h>#include<stdio.h>using namespace std;#define LL long longLL t,n,m;int main(){    scanf("%lld",&t);    while(t--)    {        scanf("%lld%lld",&n,&m);        if(m>=n*(n-1)/2)            printf("%lld\n",n*(n-1));//每个点到另一点都是距离为1        else if(m>=n-1)        {            LL more = m - n + 1;//将所有点连通后多出的边数            printf("%lld\n",2*((n-1)*(n-1) - more));  //2*(n-1)^2就是使用n-1条边将所有点连起来后所有两点距离之和        }        else        {            LL out = n - m - 1;//out表示未能连入图中的点的个数            LL ans = 2*(n-out-1)*(n-out-1)+n*out*(n-out)+n*(n-1)*out;//n-out为连通的点的个数            //ans = 连通的点之间的距离总和（双向，所以*2）+ 连通图内各点连到连通图外各点的边的总数*n（单向，从内连到外）+ 图外各点连到其余所有点的变数之和*n（单向，从图外一点出发）            printf("%lld\n",ans);        }    }    return 0;}