Hdu6090 Rikka with Graph（2017多校第5场）

Rikka with Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 683    Accepted Submission(s): 400

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
  

 

Source

2017 Multi-University Training Contest - Team 5

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题目大意：

给定n个点，m条边，让你安排点和边构成一个无向图。dis(i,j)，表示i到j最小的边数，如果无法到达，dis(i,j)为n,问每个点到其他所有的点的dis之和。

思路：

要尽可能短，肯定优先把一个点和剩下所有点相连，接下来每多一个点把叶子节点两两连起来，边不够和多出来分别特殊处理

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <unordered_map>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int T;
LL n,m;

int main()
{

    for(scanf("%d",&T); T--;)
    {
        scanf("%lld%lld",&n,&m);
        LL sum = 0;
        if(m >= n - 1)
        {
            m = min(m, n * (n - 1) / 2);
            sum += 2 * (n - 1) * (n - 1);
            m -= n - 1;
            sum -= m * 2;
        }
        else
        {
            sum += 2 * m * m;
            sum += (m + 1) * (n - m -1) * n;
            sum += (n - m -1) * n * (n - 1);
        }
        printf("%lld\n", sum);
    }
    return 0;
}