题意

对于每个非叶子节点，在子树中找到两个叶子节点使得它们的权值之差的绝对值最小。如果子树内只有一个节点。输出$2^{31}-1$

题解

启发式合并

代码

#include<bits/stdc++.h>
using namespace std;
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
const int nmax = 60000;
const int INF = 0x3f3f3f3f;
 const int thisinf = (1<<31) - 1;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const ull p = 67;
const ull MOD = 1610612741;
int n,m,temp;
int tot,ans[nmax],head[nmax],sz[nmax];
multiset<int> ms[nmax];
multiset<int> :: iterator it,pre,suc;
struct edge {int to,nxt;} e[nmax<<1];
void add_edge(int u, int v) {e[tot].to = v; e[tot].nxt = head[u]; head[u] = tot++;}
void merge(int u, int v){
    for(it = ms[v].begin();it!=ms[v].end();++it){
        pre = suc = ms[u].lower_bound(*it);
        if(pre != ms[u].begin()) pre --, temp = min(temp, abs(*pre - *it));
        if(suc != ms[u].end()) temp = min(temp, abs(*suc - * it));
        ms[u].insert(*it);
    }
}
void dfs(int u, int f) {
    ans[u] = thisinf ;
    for(int i = head[u]; i!=-1; i = e[i].nxt) {
        int v = e[i].to;
        if(v != f) {
            dfs(v,u);
            temp = thisinf;
            ans[u] = min(ans[u],ans[v]);
            if(ms[u].size() <= ms[v].size()) merge(v,u), swap(ms[u],ms[v]);
            else merge(u,v);
            ans[u] = min(ans[u],temp);
        }
    }
}
int main(){
    memset(head,-1,sizeof head);
    scanf("%d %d",&n,&m);
    int v;
    for(int u = 2;u<=n;++u) scanf("%d",&v),add_edge(u,v), add_edge(v,u);
    for(int i = n-m+1;i<=n;++i)  scanf("%d",&temp), ms[i].insert(temp);
    dfs(1,-1);
    for(int i = 1;i<=n-m;++i) printf("%d ",ans[i]);
    return 0;
}