hdu 6090（二）

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Rikka with Graph

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 72 Accepted Submission(s): 51

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

题目大意：给你两个数字n，m分别代表的是有多少点多少边，每个边的长度为1，现在让你求m条边构成的图中，所有的点相连所需要的长度和最小。

解题思路：找规律，发现m在不同的范围内，所求的答案有不同的公式表示，每两个点之间共有n*（n-1）/2种可能如果m大于等于这个数就直接为n*（n-1）如果少于它则一个边一个边的进行拆除没拆一个边就增加2然后到一个临界点为m=n-1一个点分别连其他的点可以根据上一个算出，当m小于这个数的时候就分为孤立的点了，有一部分相连，就和上一种情况一样然后分联通的，孤立的，联通的与孤立的分别计算。

[cpp] view plain copy

print?

#include<cstdio>
using namespace std;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
long long n,m,p,q;
scanf("%lld%lld",&n,&m);
long long ans=0;
int flag=0;
if(m>n*(n-1)/2)///在这个范围内代表每两个点之间都可以一步连到所以为n*（n-1）
{
printf("%lld\n",n*(n-1));
flag=1;
}
else if(m>=n-1&&m<=n*(n-1)/2)///在这个范围内与m为n*（n-1）/2相比每少一条边就减少2
{
ans+=((n*(n-1)/2)-m)*2+n*(n-1);
}
else///这个范围内分为孤立的点与连在一起的点，分三块计算
{
p=m+1;
q=n-m-1;
ans+=q*p*n*2+q*(q-1)*n+(p-1)*(p-1)*2;
}
if(flag==0)
printf("%lld\n",ans);
}
}

#include<cstdio>using namespace std;int main(){    int t;    scanf("%d",&t);    while(t--)    {        long long n,m,p,q;        scanf("%lld%lld",&n,&m);        long long ans=0;        int flag=0;        if(m>n*(n-1)/2)///在这个范围内代表每两个点之间都可以一步连到所以为n*（n-1）        {            printf("%lld\n",n*(n-1));            flag=1;        }        else if(m>=n-1&&m<=n*(n-1)/2)///在这个范围内与m为n*（n-1）/2相比每少一条边就减少2      {            ans+=((n*(n-1)/2)-m)*2+n*(n-1);        }        else///这个范围内分为孤立的点与连在一起的点，分三块计算        {            p=m+1;            q=n-m-1;            ans+=q*p*n*2+q*(q-1)*n+(p-1)*(p-1)*2;        }        if(flag==0)        printf("%lld\n",ans);    }}

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