给你一颗树(可能有好几棵),你每次最多只能去掉k个叶子节点,问你最多几次能去完。
按深度deep保存每个深度的节点个数,从前往后for一遍,过程中如果一个deep大于k,就往后撩,最后到0为止,长度就是次数。(队友想出来的,我没这么聪明??)
1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0); 2 #include <cstdio>//sprintf islower isupper 3 #include <cstdlib>//malloc exit strcat itoa system("cls") 4 #include <iostream>//pair 5 #include <fstream>//freopen("C:\\Users\\13606\\Desktop\\Input.txt","r",stdin); 6 #include <bitset> 7 //#include <map> 8 //#include<unordered_map> 9 #include <vector> 10 #include <stack> 11 #include <set> 12 #include <string.h>//strstr substr strcat 13 #include <string> 14 #include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9; 15 #include <cmath> 16 #include <deque> 17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less 18 #include <vector>//emplace_back 19 //#include <math.h> 20 #include <cassert> 21 #include <iomanip> 22 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor 23 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare) 24 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation 25 //****************** 26 clock_t __STRAT,__END; 27 double __TOTALTIME; 28 void _MS(){__STRAT=clock();} 29 void _ME(){__END=clock();__TOTALTIME=(double)(__END-__STRAT)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;} 30 //*********************** 31 #define rint register int 32 #define fo(a,b,c) for(rint a=b;a<=c;++a) 33 #define fr(a,b,c) for(rint a=b;a>=c;--a) 34 #define mem(a,b) memset(a,b,sizeof(a)) 35 #define pr printf 36 #define sc scanf 37 #define ls rt<<1 38 #define rs rt<<1|1 39 typedef pair<int,int> PII; 40 typedef vector<int> VI; 41 typedef long long ll; 42 const double E=2.718281828; 43 const double PI=acos(-1.0); 44 const ll INF=(1LL<<60); 45 const int inf=(1<<30); 46 const double ESP=1e-9; 47 const int mod=(int)1e9+7; 48 const int N=(int)1e6+10; 49 50 vector<vector<int> >G(N); 51 int deep[N]; 52 53 struct node 54 { 55 int x,dep; 56 }; 57 bool vis[N]; 58 void bfs(int start) 59 { 60 queue<node>q; 61 q.push({start,1}); 62 while(!q.empty()) 63 { 64 node now=q.front();q.pop(); 65 deep[now.dep]++; 66 vis[now.x]=1; 67 int sz=G[now.x].size(); 68 for(int i=0;i<sz;++i) 69 { 70 int to=G[now.x][i]; 71 if(!vis[to]) 72 q.push({to,now.dep+1}); 73 } 74 } 75 } 76 77 int a[N]; 78 int main() 79 { 80 int n,k; 81 sc("%d%d",&n,&k); 82 for(int i=1;i<=n;++i) 83 { 84 sc("%d",&a[i]); 85 if(a[i]==0)continue; 86 G[a[i]].push_back(i); 87 G[i].push_back(a[i]); 88 } 89 for(int i=1;i<=n;++i) 90 if(a[i]==0) 91 bfs(i); 92 for(int i=1;i<=n+5;++i) 93 { 94 if(deep[i]==0) 95 { 96 pr("%d\n",i-1); 97 break; 98 } 99 if(deep[i]>k) 100 deep[i+1]+=deep[i]-k; 101 } 102 return 0; 103 } 104 105 /**************************************************************************************/
原文:https://www.cnblogs.com/--HPY-7m/p/11873038.html