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  • CF600E Lomsat gelral

    // Problem: CF600E Lomsat gelral
    // Contest: Luogu
    // URL: https://www.luogu.com.cn/problem/CF600E
    // Memory Limit: 250 MB
    // Time Limit: 2000 ms
    // 
    // Powered by CP Editor (https://cpeditor.org)
    #include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    typedef unsigned long long ull;
    typedef pair<ll, ll>PLL;
    typedef pair<int, int>PII;
    typedef pair<double, double>PDD;
    #define I_int ll
    inline ll read()
    {
        ll x = 0, f = 1;
        char ch = getchar();
        while(ch < '0' || ch > '9')
        {
            if(ch == '-')f = -1;
            ch = getchar();
        }
        while(ch >= '0' && ch <= '9')
        {
            x = x * 10 + ch - '0';
            ch = getchar();
        }
        return x * f;
    }
    
    inline void out(ll x){
    	if (x < 0) x = ~x + 1, putchar('-');
    	if (x > 9) out(x / 10);
    	putchar(x % 10 + '0');
    }
    
    inline void write(ll x){
    	if (x < 0) x = ~x + 1, putchar('-');
    	if (x > 9) write(x / 10);
    	putchar(x % 10 + '0');
    	puts("");
    }
    
    #define read read()
    #define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
    #define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
    #define rep(i,a,b) for(int i=(a);i<=(b);i++)
    #define repp(i,a,b) for(int i=(a);i<(b);i++)
    #define per(i,a,b) for(int i=(a);i>=(b);i--)
    #define perr(i,a,b) for(int i=(a);i>(b);i--)
    ll ksm(ll a, ll b, ll p)
    {
        ll res = 1;
        while(b)
        {
            if(b & 1)res = res * a % p;
            a = a * a % p;
            b >>= 1;
        }
        return res;
    }
    const int inf = 0x3f3f3f3f;
    #define PI acos(-1)
    const int maxn=100000+100;
    
    int n,col[maxn];
    vector<int>g[maxn];
    ll ans[maxn],sum;
    int son[maxn],siz[maxn],cnt[maxn],maxx,Son;
    
    void dfs1(int u,int fa){
    	siz[u]=1;
    	for(int i=0;i<g[u].size();i++){
    		int j=g[u][i];
    		if(j==fa) continue;
    		dfs1(j,u);
    		siz[u]+=siz[j];
    		if(siz[j]>siz[son[u]]) son[u]=j;
    	}
    }
    
    void add(int u,int fa,int val){
    	cnt[col[u]]+=val;
    	if(cnt[col[u]]>maxx) maxx=cnt[col[u]],sum=col[u];
    	else if(cnt[col[u]]==maxx) sum+=col[u]*1ll;
    	for(int i=0;i<g[u].size();i++){
    		int j=g[u][i];
    		if(j==fa||j==Son) continue;
    		add(j,u,val);
    	}
    }
    
    void dfs2(int u,int fa,int op){
    	for(int i=0;i<g[u].size();i++){
    		int j=g[u][i];
    		if(j==fa) continue;
    		if(j!=son[u]) dfs2(j,u,0);
    	}
    	if(son[u]) dfs2(son[u],u,1),Son=son[u];
    	add(u,fa,1);Son=0;
    	ans[u]=sum;
    	if(!op) add(u,fa,-1),sum=0,maxx=0;
    }
    
    int main(){
    	n=read;
    	rep(i,1,n) col[i]=read;
    	rep(i,1,n-1){
    		int u=read,v=read;
    		g[u].push_back(v);
    		g[v].push_back(u);
    	}	
    	dfs1(1,0);
    	dfs2(1,0,0);
    	rep(i,1,n) printf("%lld ",ans[i]);
    	
    	
    	return 0;
    } 
    
    
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  • 原文地址:https://www.cnblogs.com/OvOq/p/15049585.html
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