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
题意：求在一个有根节点的树中 每个节点的子树中有多少个小于改点的序号
DFS序找到每个节点的时间戳 用树状数组直接维护这个区间就可以了
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 }