HDU 3887 Counting Offspring(dfs序 + 树状数组)

Counting Offspring

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2201    Accepted Submission(s): 738


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

解题思路：

求每个节点的字数中，标号比它小的点的个数，通过dfs转化为序列，然后用树状数组求解。与上一题类似，不过dfs过程中保存的东西不一样，本题保存的是所有点的标号，上一题保存的是每个点的进入时间和离开时间。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <vector>
#include <cstring>
using namespace std;
const int MAXN = 100000 + 10;
struct Edge{int to, next;}edge[MAXN<<1];
int tot, head[MAXN], n, root, dfs_clock, p[MAXN<<1], vis[MAXN],C[MAXN<<1], u, v, f[MAXN];
int H[MAXN];
void init()
{
tot = 0; memset(head, -1, sizeof(head));
dfs_clock = 0; memset(vis, 0, sizeof(vis));
memset(C, 0, sizeof(C));
}
void addedge(int u, int v)
{
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
}
void dfs(int x)
{
if(vis[x]) return ;
vis[x] = 1; p[dfs_clock++] = x;
for(int i=head[x];i!=-1;i=edge[i].next)
{
int v = edge[i].to;
if(!vis[v]) dfs(v);
}
p[dfs_clock++] = x;
}
void add(int x, int d)
{
while(x <= 2 * n){C[x] += d; x += x & -x;}
}
int sum(int x)
{
int rs = 0;
while(x){rs += C[x]; x -= x & -x;}
return rs;
}
int main()
{
while(scanf("%d%d", &n, &root)!=EOF)
{
if(n == 0 && root == 0) break;
init();
for(int i=1;i<n;i++)
{
scanf("%d%d", &u, &v);
addedge(u, v); addedge(v, u);
}
dfs(root);
for(int i=0;i<=n;i++) H[i] = -1;
for(int i=0;i<dfs_clock;i++)
{
if(H[p[i]] != -1) f[p[i]] = sum(p[i] - 1) - H[p[i]];
else
{
H[p[i]] = sum(p[i] - 1);
add(p[i], 1);
}
}
for(int i=1;i<n;i++) cout << f[i] << ' ';
cout << f
<< endl;
}
return 0;
}
/*
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
*/