Hdu 6090 Rikka with Graph【贪心】

Rikka with Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 112    Accepted Submission(s): 81

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

题目大意：

一个有N个点的完全无向图，希望我们从中选出不超过M条边，使得选出的图 ∑ni=1∑nj=1dist(i,j).    的值最小。

思路：

显然类似这样的菊花图去取边是最优的方案。

那么考虑：

①最初如果M=0的时候，答案是n^3-n^2

②当M=1的时候，对答案减少的贡献是n-1.

③当M>1&&M<=n-1的时候，每条边对答案减少的贡献是(i-1)*(n-2)+(n-1)；

④当M>n-1的时候，每条边对答案减少的贡献是2或者是0.

⑤答案最小是n^2-n；

那么我们暴力搞一下就行了。

Ac代码：

#include<stdio.h>#include<iostream>#include<string.h>#include<algorithm>using namespace std;#define ll __int64int a[1500000];int main(){    int t;    scanf("%d",&t);    while(t--)    {        ll n,m;        scanf("%I64d%I64d",&n,&m);        ll Ans=n*n*n-n*n;        for(ll i=1;i<=n-1&&i<=m;i++)        {            if(i==1)Ans-=2*(n-1);            else Ans-=2ll*((i-1)*(n-2)+(n-1));        }        m=max(0ll,m-(n-1));        Ans-=2*m;        Ans=max(Ans,n*n-n);        printf("%I64d\n",Ans);    }}