标签:queue clu efi test sync tac namespace com cout
题意:构造一颗n个节点k个叶子节点是树,使得叶子节点的最远距离最小
思路:贪心构造,先构造一个k+1个节点,k个叶子节点的树,用数组标记每个叶子节点的编号,然后剩下n-k-1个节点依次往每个叶子节点上加,并更新叶子节点
AC代码:
#include "iostream" #include "iomanip" #include "string.h" #include "stack" #include "queue" #include "string" #include "vector" #include "set" #include "map" #include "algorithm" #include "stdio.h" #include "math.h" #pragma comment(linker, "/STACK:102400000,102400000") #define bug(x) cout<<x<<" "<<"UUUUU"<<endl; #define mem(a,x) memset(a,x,sizeof(a)) #define step(x) fixed<< setprecision(x)<< #define mp(x,y) make_pair(x,y) #define pb(x) push_back(x) #define ll long long #define endl ("\n") #define ft first #define sd second #define lrt (rt<<1) #define rrt (rt<<1|1) using namespace std; const ll mod=1e9+7; const ll INF = 1e18+1LL; const int inf = 1e9+1e8; const double PI=acos(-1.0); const int N=2e5+100; int main(){ ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); int n,k; cin>>n>>k; int t=(n-1)/k, ans=t,x[N]; if(n-(t*k+1)==0) ans+=t; else if(n-(t*k+1)==1) ans+=t+1; else ans+=t+2; cout<<ans<<endl; for(int i=1; i<=k; ++i){ cout<<i<<" "<<k+1<<endl; x[i]=i; } int now=0; for(int i=k+2; i<=n; ++i){ now%=k; cout<<x[now+1]<<" "<<i<<endl; x[++now]=i; } return 0; }
标签:queue clu efi test sync tac namespace com cout
原文地址:http://www.cnblogs.com/max88888888/p/7554768.html