題目連結

題意

構造一顆樹使得滿足計算方法的結果最小。

思路

考慮兩棵樹，一棵為題目中的詢問構成的樹T1，一棵為要構造的滿足最終答案的樹T2。從T1點權最小的點向外構造T2，在T1中倍增預處理出祖先，每次選取結果最小的作為對答案的貢獻，在T2中讓當前點連上這個祖先。由於點權最小的點為根，所以構造的過程中當前點始終滿足點權大於父節點。

程式碼

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <vector>
#include <cassert>
#include <cstring>
#include <iostream>
#include <algorithm>

#define IOS    ios::sync_with_stdio(0),cin.tie(0);
#define DBG(x) cerr << #x << " = " << x << endl;

using namespace std;

typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;

const int inf = 0x3f3f3f3f;
const int mod = 1000000007;
const double eps = 1e-8;
const double pi  = acos(-1.0);

void file(){
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
}

const int maxn = 5e5+5;
const int maxm = 1e6+5;

int n;
int fa[maxn][25];
int tot,head[maxn];
LL minn=inf,pos,ans;
LL a[maxn];

struct edgenode{
    int to,next;
}edge[maxm];

void addedge(int u,int v){
    edge[tot].to=v;
    edge[tot].next=head[u];
    head[u]=tot++;
}

void dfs(int x,int pre){
    fa[x][0]=pre;
    for(int i=1;i<=20;i++)fa[x][i]=fa[fa[x][i-1]][i-1];
    if(x != pre){
        LL tmp=a[pre];
        for(int i=1;i<=20;i++)tmp=min(tmp,a[fa[x][i]]*1LL*(i+1));
        ans+=a[x]+tmp;
    }
    for(int i=head[x];i != -1;i=edge[i].next){
        int v=edge[i].to;
        if(v != pre)dfs(v,x);
    }
}

namespace BakuretsuMahou{
    void Explosion(){
        memset(head,-1,sizeof head);
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%I64d",&a[i]);
            if(a[i] < minn){
                minn=a[i];
                pos=i;
            }
        }
        for(int i=1,x,y;i<=n-1;i++){
            scanf("%d%d",&x,&y);
            addedge(x,y);
            addedge(y,x);
        }
        dfs(pos,pos);
        printf("%I64d\n",ans);
    }
}

int main(){
    //IOS
    //file();
    BakuretsuMahou::Explosion();
    return 0;
}