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  • https://vjudge.net/problem/计蒜客-44317/origin

    让最大值走得最多就好了,让最大值走最长链的长度,然后把最长链上的点都打上标记,再从所有没有被标记的点里找次长链,让次大值跑
    这个题我最后一小时写慌了,思路乱了,队友写得很妙,tql

    #include <iostream>
    #include <cstdio>
    #include <queue>
    #include <algorithm>
    #include <cmath>
    #include <cstring>
    #define inf 2147483647
    #define N 1000010
    #define p(a) putchar(a)
    #define For(i,a,b) for(long long i=a;i<=b;++i)
    
    using namespace std;
    long long n,k,ans,cnt,f,x,s;
    long long a[N],d[N],son[N];
    bool vis[N];
    struct tree{
        long long id;
        long long dep;
        bool operator < (const tree &x) const{
            return dep>x.dep;
        }
    }b[N];
    
    struct node{
        long long n;
        node *next;
    }*e[N];
    
    void in(long long &x){
        long long y=1;char c=getchar();x=0;
        while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
        while(c<='9'&&c>='0'){ x=(x<<1)+(x<<3)+c-'0';c=getchar();}
        x*=y;
    }
    void o(long long x){
        if(x<0){p('-');x=-x;}
        if(x>9)o(x/10);
        p(x%10+'0');
    }
    
    void push(long long x,long long y){
        node *p;
        p=new node();
        p->n=y;
        if(e[x]==0)
            e[x]=p;
        else{
            p->next=e[x]->next;
            e[x]->next=p;
        }
    }
    
    void dfs(long long x){
        b[x].dep=1;
        for(node *i=e[x];i;i=i->next){
            if(i->n!=d[x]){
                dfs(i->n);
                if(b[i->n].dep+1>b[x].dep){
                    b[x].dep=b[i->n].dep+1;
                    son[x]=i->n;
                }
            }
        }
    }
    
    bool cmp(long long x,long long y){
        return x>y;
    }
    
    signed main(){
        in(n);in(k);d[1]=1;
        For(i,1,k) in(a[i]);
        For(i,2,n){
            in(x);
            d[i]=++x;
            push(i,x);push(x,i);
        }
        For(i,1,n) b[i].id=i;
        dfs(1);
        sort(a+1,a+n+1,cmp);
        sort(b+1,b+n+1);
        b[1].dep--;
        For(i,1,n){
            if(!vis[b[i].id]){
                vis[b[i].id]=1;
                ans+=a[++cnt]*b[i].dep;
                s=son[b[i].id];
                while(s){
                    vis[s]=1;
                    s=son[s];
                }
            }
        }
        o(ans);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/war1111/p/12569245.html
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