Rikka with Graph
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 679 Accepted Submission(s): 397
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 graphG 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 graphG (wG )
as ∑ni=1∑nj=1dist(i,j) .
Now, Yuta hasn 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 ofwG .
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) .
For an undirected graph
Then, we can define the weight of the graph
Now, Yuta has
Yuta wants to know the minimal value of
It is too difficult for Rikka. Can you help her?
In the sample, Yuta can choose
Input
The first line contains a number t(1≤t≤10) ,
the number of the testcases.
For each testcase, the first line contains two numbersn,m(1≤n≤106,1≤m≤1012) .
For each testcase, the first line contains two numbers
Output
For each testcase, print a single line with a single number -- the answer.
Sample Input
1 4 5
Sample Output
14
Source
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <string> #include <map> #include <cstring> #define INF 0x3f3f3f3f #define ms(x,y) memset(x,y,sizeof(x)) using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int, int> P; const int maxn = 30010; const int mod = 998244353; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 int main() { std::ios::sync_with_stdio(false); int t; cin >> t; while (t--) { ll n, m, ans = 0; cin >> n >> m; if (m <= n - 1) { ll p = m + 1, q = n - p; //p连接的点,q孤立的点 ans = (2 * p - 2)*(p - 1) + q*(q - 1)*n + 2 * p*q*n; } else if (m <= (n - 1)*n / 2) { m = (n - 1)*n / 2 - m; ans = n*(n - 1); ans += 2 * m; } else { ans = n*(n - 1); } cout << ans << endl; } return 0; }