pac ria div cst key rst n-1 include http

Counting Offspring

Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3054 Accepted Submission(s): 1031

Problem Description You are given a tree, it’s root is p, and the node is numbered from 1 to n. Now define f(i) as the number of nodes whose number is less than i in all the succeeding nodes of node i. Now we need to calculate f(i) for any possible i. Input Multiple cases (no more than 10), for each case:
The first line contains two integers n (0<n<=10^5) and p, representing this tree has n nodes, its root is p.
Following n-1 lines, each line has two integers, representing an edge in this tree.
The input terminates with two zeros. Output For each test case, output n integer in one line representing f(1), f(2) … f(n), separated by a space. Sample Input 15 7 7 10 7 1 7 9 7 3 7 4 10 14 14 2 14 13 9 11 9 6 6 5 6 8 3 15 3 12 0 0 Sample Output 0 0 0 0 0 1 6 0 3 1 0 0 0 2 0 Author bnugong Source 2011 Multi-University Training Contest 5 - Host by BNU 題意：求在一個有根節點的樹中 每個節點的子樹中有多少個小於改點的序號 DFS序找到每個節點的時間戳 用樹狀數組直接維護這個區間就可以了

 1
 
 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cctype>
 5 #include<cmath>
 6 #include<cstring>
 7 #include<map>
 8 #include<set>
 9 #include<queue>
10 #include<vector>
11 #include<algorithm>
12 #include<string
 
>
13 #define ll long long
14 #define eps 1e-10
15 #define LL unsigned long long
16 using namespace std;
17 const int inf=0x3f3f3f3f;
18 const int N=200000+10;
19 const int mod=1e9+7;
20 int head[N];
21 int tot,time;
22 int L[N],R[N];
23 struct node{
24     int to,next;
25 }edge[N<<1];
26 int a[N*5];
27 void init(){
28     memset(head,-1,sizeof(head));
29     tot=0;
30     time=0;
31 }
32 void add(int u,int v){
33     edge[tot].to=v;
34     edge[tot].next=head[u];
35     head[u]=tot++;
36 }
37 void DFS(int x,int fa){
38     L[x]=++time;
39     for(int i=head[x];i!=-1;i=edge[i].next){
40         int v=edge[i].to;
41         if(v==fa)continue;
42         DFS(v,x);
43     }
44     R[x]=time;
45 
46 }
47 int lowbit(int x){
48     return x&(-x);
49 }
50 void update(int x,int y){
51     while(x<=time){
52         a[x]=a[x]+y;
53         x=x+lowbit(x);
54     }
55 }
56 int getsum(int x){
57     int ans=0;
58     while(x>0){
59         //cout<<2<<endl;
60         ans=ans+a[x];
61         x=x-lowbit(x);
62     }
63     return ans;
64 }
65 int main(){
66     int n,p;
67     while(scanf("%d%d",&n,&p)!=EOF){
68         if(n==0&&p==0)break;
69         init();
70         memset(a,0,sizeof(a));
71         for(int i=1;i<n;i++){
72             int u,v;
73             scanf("%d%d",&u,&v);
74             add(u,v);
75             add(v,u);
76         }
77         DFS(p,-1);
78         for(int i=1;i<=n;i++){
79             int ans=getsum(R[i])-getsum(L[i]-1);
80             if(i!=n)cout<<ans<<" ";
81             else{
82                 cout<<ans<<endl;
83             }
84             update(L[i],1);
85         }
86     }
87 }

HDU 3887 Counting Offspring（DFS序）