HDU 6090 Rikka with Graph（思维）_touch it rikka-优快云博客

本文介绍了一道图论问题，任务是在给定的节点数和边数条件下，构造一张图使得所有节点对之间的最短路径之和最小。文章提供了问题背景、输入输出说明及样例解释，并给出了一个具体的解题思路与代码实现。

Rikka with Graph

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

Source

2017 Multi-University Training Contest - Team 5

题意：给出n个孤立点，使用m条边连接，使得任意两点间距离之和最短（距离等于经过的边数，不连通的两点距离为n）

题解：考虑贪心地一条一条边添加进去。

当 $\leq n-1$ 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 $\frac{n(n-1)}{2}$ 。

#include<iostream>
#include<stdio.h>
#define ll long long
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        ll n,m;
        scanf("%lld%lld",&n,&m);
        ll ans=0;
        if(m<=n-1)
        {
            ll s=n-1-m;
            ans=m*2+(m-1)*m*2+s*(m+1)*n*2+s*(s-1)*n;
        }
        else if(m<n*(n-1))
        {
            ll s=m-n+1;
            ans=(n-1)*2+2*(n-1)*(n-2)-2*s;
        }
        else ans=n*(n-1);
        cout<<ans<<endl;
    }
    return  0;
}