标签：ace this present lines pre oss term wing ack
[Hdu3887]Counting Offspring
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.
给你一个树，问你对于每个节点，它的子树上标号比它小的点有多少个
N<=100000
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
Sol:跑出dfs序来，然后树状数组查询下就好了。
#include<cstdio>
#include<vector>
#include<cstring>
#include<bits/stdc++.h>
using namespace std;
const int maxn = 100000+10;
vector<int>G[maxn];
int n,p,t;
int l[maxn],r[maxn],tree[maxn];
void dfs(int u,int fa)
//求出dfs序，某个树上所有结点必然是连在一起的
{
l[u] = ++t;
int len = G[u].size();
for(int i=0;i<len;i++)
{
int v = G[u][i];
if(v==fa) continue;
dfs(v,u);
}
r[u] = t;
}
inline int lowbit(int i)
{
return i&(-i);
}
inline void add(int i,int d)
{
while(i<=n)
{
tree[i] += d;
i += lowbit(i);
}
}
inline int getsum(int i)
{
int ans = 0;
while(i)
{
ans += tree[i];
i -= lowbit(i);
}
return ans;
}
int main()
{
int u,v;
while(scanf("%d%d",&n,&p))//N个点,P为root点
{
if(n==0&&p==0) break;
memset(tree,0,sizeof(tree));
for(int i=0;i<=n;i++)
G[i].clear();
for(int i=1;i<n;i++)
{
scanf("%d%d",&u,&v);
G[u].push_back(v);
G[v].push_back(u);
}
t = 0;
dfs(p,-1);
// for (int i=1;i<=n;i++)
//cout<<i<<" "<<l[i]<<" "<<r[i]<<endl;
for(int i=1;i<=n;i++)
//统计对于i来说，有多少个点值比它小
{
printf("%d",getsum(r[i])-getsum(l[i]-1));
//统计从[l(i),r(i)]这一段中有多少数字比i小
if(i==n)
printf("\n");
else
printf(" ");
add(l[i],1);
}
}
return 0;
}
/*
input
4 4
4 3
4 2
3 1
结点编号 入 出
1 3 3
2 4 4
3 2 3
4 1 4
output
0 0 1 3
*/
标签：ace this present lines pre oss term wing ack
原文地址：https://www.cnblogs.com/cutemush/p/11794548.html
