传送门
题目描述
给定一棵树,树根的编号为p。
对于每个节点,求以它为根的子树中有多少个节点的编号小于它。
分析
我们可以根据节点的大小,然后把对应位置设置为1,然后我们要求某一个节点的答案的时候,只需要求子树的值即可
我们可以用dfs序和树状数组去维护答案,最后求和即可
代码
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <queue>
#include <cstring>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define _CRT_SECURE_NO_WARNINGS
#pragma GCC optimize("Ofast","unroll-loops","omit-frame-pointer","inline")
#pragma GCC option("arch=native","tune=native","no-zero-upper")
#pragma GCC target("avx2")
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
const int INF = 0x3f3f3f3f;
const int N = 1e5 + 10,M = 2 * N;
int n,p;
int h[N],ne[M],e[M],idx;
int in[N],out[N];
int cnt;
int tr[N];
int lowbit(int x){
return x & -x;
}
void add(int x,int y){
ne[idx] = h[x],e[idx] = y,h[x] = idx++;
}
void ad(int x,int c){
for(int i = x;i <= n;i += lowbit(i)) tr[i] += c;
}
int sum(int x){
int res = 0;
for(int i = x;i;i -= lowbit(i)) res += tr[i];
return res;
}
void dfs(int u,int fa){
in[u] = ++cnt;
for(int i = h[u];~i;i = ne[i]){
int j = e[i];
if(j == fa) continue;
dfs(j,u);
}
out[u] = cnt;
}
int main(){
while(scanf("%d%d",&n,&p) && n && p){
memset(h,-1,sizeof h);
idx = cnt = 0;
for(int i = 1;i < n;i++){
int x,y;
scanf("%d%d",&x,&y);
add(x,y),add(y,x);
}
dfs(p,-1);
memset(tr,0,sizeof tr);
for(int i = 1;i <= n;i++){
int ans = sum(out[i]) - sum(in[i]);
if(i != n) printf("%d ",ans);
else printf("%d\n",ans);
ad(in[i],1);
}
}
}
/**
* ┏┓ ┏┓+ +
* ┏┛┻━━━┛┻┓ + +
* ┃ ┃
* ┃ ━ ┃ ++ + + +
* ████━████+
* ◥██◤ ◥██◤ +
* ┃ ┻ ┃
* ┃ ┃ + +
* ┗━┓ ┏━┛
* ┃ ┃ + + + +Code is far away from
* ┃ ┃ + bug with the animal protecting
* ┃ ┗━━━┓ 神兽保佑,代码无bug
* ┃ ┣┓
* ┃ ┏┛
* ┗┓┓┏━┳┓┏┛ + + + +
* ┃┫┫ ┃┫┫
* ┗┻┛ ┗┻┛+ + + +
*/