Rikka with Graph （2017多校图论规律题）

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本文探讨了一个关于图论的问题RikkawithGraph，目标是在给定节点数量和最大边数的情况下，构建一个无向图并求出节点间最短路径之和的最小值。文章给出了具体的算法实现，并通过样例进行了解释。

Rikka with Graph

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 234 Accepted Submission(s): 154

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

Source

2017 Multi-University Training Contest - Team 5

我是从星状图画了一下找到了规律最小的结果肯定是完全图的时候边的个数大于等于（n * （n - 1））/ 2的时候。然后随便推推就出来了

#include <iostream>
#include <string>
#include <string.h>
#include <algorithm>
#include <cmath>
#include <vector>
#include <stdio.h>
#include <queue>
using namespace std;
const int maxn = 50005;
int main() {
        // freopen("1.txt", "r", stdin);
    long long n, m;
    int t;
    cin >> t;
    while (t--) {
        cin >> n >> m;
        long long up = (n * (n - 1)) / 2;
        if (m >= up) {
            cout << up * 2 << endl;
        } else if (m >= n - 1) {
            long long lose = up - m;
                //cout << lose << endl;
            cout << (up + lose) * 2 << endl;
        } else {
            long long a = m * 2 * m  + (n - m - 1) * n * (m + 1);
            long long b = (n - m - 1) * (n - 1) * n;
            cout << a + b << endl;
        }
    }
}