2017多校训练Contest5: 1006 Rikka with Graph hdu6090

最新推荐文章于 2017-09-05 15:10:55 发布

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最新推荐文章于 2017-09-05 15:10:55 发布

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本文介绍了一种算法，用于解决如何构建一个具有最小权重的无向图的问题。该图包含n个节点和不超过m条边，通过精心选择连接哪些节点来最小化所有节点对之间的最短路径总和。

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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

Sample Output

考虑直接以一个点为中心练成一个放射状的图

多出来的边再把发散出来的点两两连线

维护下每种情况的贡献即可

#include<map>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<cassert>
#include<iostream>
#include<algorithm>
using namespace std;
int a[100001];
int main()
{
    int T;
    scanf("%d",&T);
    while(T>0)
    {
        T--;
        long long n,m;
        scanf("%I64d%I64d",&n,&m);
        if(m>=n*(n-1LL)/2LL)
        {
            printf("%I64d\n",n*(n-1LL));
            continue;
        }
        long long ans=(n-1LL)*n*n;
        long long i;
        if(m<=n-1LL)
        {
            long long d=0;
            for(i=1;i<=m;i++)
            {
                ans-=2LL*((n-1LL)+d*(n-2LL));
                d++;
            }
        }
        else
        {
            long long d=0;
            for(i=1;i<=n-1LL;i++)
            {
                ans-=2LL*((n-1LL)+d*(n-2LL));
                d++;
            }
            m-=(n-1LL);
            ans-=2LL*m;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}