HDU6090-Rikka with Graph

Rikka with Graph

                                                                     Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                               Total Submission(s): 794    Accepted Submission(s): 452

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

 

题意：给出n个点，在它们之间加入m条边，使得任意两点的距离和最小（两点距离为两点之间的最短路，一条边的距离为1，两点之间不能达到则距离为n）

解题思路：先随意找一个根节点，在m足够大的条件下，将它和其他所有点相连，则此时答案为2*（n-1）*（n-1），若边有多余，则在除根节点的任意两点之间加边，每加一条边答案减2

#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 longconst int INF = 0x3f3f3f3f;LL n,m;int main(){    int t;    scanf("%d", &t);    while(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;}