Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals) D. High Load 构造

D. High Load

题目连接:

http://codeforces.com/contest/828/problem/D

Description

Arkady needs your help again! This time he decided to build his own high-speed Internet exchange point. It should consist of n nodes connected with minimum possible number of wires into one network (a wire directly connects two nodes). Exactly k of the nodes should be exit-nodes, that means that each of them should be connected to exactly one other node of the network, while all other nodes should be connected to at least two nodes in order to increase the system stability.

Arkady wants to make the system as fast as possible, so he wants to minimize the maximum distance between two exit-nodes. The distance between two nodes is the number of wires a package needs to go through between those two nodes.

Help Arkady to find such a way to build the network that the distance between the two most distant exit-nodes is as small as possible.

Input

The first line contains two integers n and k (3 ≤ n ≤ 2·105, 2 ≤ k ≤ n - 1) — the total number of nodes and the number of exit-nodes.

Note that it is always possible to build at least one network with n nodes and k exit-nodes within the given constraints.

Output

In the first line print the minimum possible distance between the two most distant exit-nodes. In each of the next n - 1 lines print two integers: the ids of the nodes connected by a wire. The description of each wire should be printed exactly once. You can print wires and wires' ends in arbitrary order. The nodes should be numbered from 1 to n. Exit-nodes can have any ids.

If there are multiple answers, print any of them.

Sample Input

3 2

Sample Output

2
1 2
2 3

Hint

题意

你需要构造n个点一棵树,满足这个树里面只有k个点的度数为1,且树的直径最小。

题解:

纸上随便画一画就可以知道,我们肯定是按照那k个平均去画就好了,也就是首先让一个节点当中央节点,然后剩下的每一个点都分(n-1)/k节点,如果n-k%k!=0的话,我们让一个节点后面增加一个就好了。 >1同理

代码

#include<bits/stdc++.h>
using namespace std;

int n,k;
vector<pair<int,int> >ans;
int main(){
    scanf("%d%d",&n,&k);
    n=n-1;
    int Ans=n/k*2+(n%k>0)+(n%k>1);
    int p=0;
    for(int i=0;i<n-n%k;i++){
        if(i%(n/k)==0)
            ans.push_back(make_pair(i+1,n+1));
        else
            ans.push_back(make_pair(i,i+1));
        if(i%(n/k)==0)p=p+1;
    }
    int step =n/k-1;
    for(int i=n-n%k;i<n;i++){
    //  cout<<p<<endl;
        if(p<k){
            p=p+1;
            ans.push_back(make_pair(step,i+1));
        }else{
            ans.push_back(make_pair(step+1,i+1));
        }
        step+=n/k;
    }
    cout<<Ans<<endl;
    sort(ans.begin(),ans.end());
    for(int i=0;i<ans.size();i++){
        cout<<ans[i].first<<" "<<ans[i].second<<endl;
    }

}
posted @ 2017-07-16 11:08 qscqesze 阅读(...) 评论(...) 编辑 收藏