• CF EDU


    学习:http://codeforces.com/blog/entry/44351

    E. Lomsat gelral 

    题意:

    给定一个以1为根节点的树,每个节点都有一个颜色,问每个节点的子树中,颜色最多的是哪几种颜色,输出这些颜色的值得和。

    思路:

    树上启发式合并的模板题,具体来说,先对树进行树链剖分,分出一个重链和轻边,dfs时,把每条轻儿子暴力加到根节点中,每次加的时候,用这样的技巧

    if(csz < cnt[col[v]]) sum = col[v] , csz = cnt[col[v]];
    else if(csz==cnt[col[v]]) sum += col[v];

    就可以使得sum中保存的是这个节点的子树中数量相同且最多的颜色和。

    #include <iostream>
    #include <cstdio>
    #include <algorithm>
    #include <cstring>
    #include <string>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <list>
    #include <cstdlib>
    #include <iterator>
    #include <cmath>
    #include <iomanip>
    #include <bitset>
    #include <cctype>
    using namespace std;
    //#pragma comment(linker, "/STACK:102400000,102400000")  //c++
    #define lson (l , mid , rt << 1)
    #define rson (mid + 1 , r , rt << 1 | 1)
    #define debug(x) cerr << #x << " = " << x << "
    ";
    #define pb push_back
    #define pq priority_queue
    
    typedef long long ll;
    typedef unsigned long long ull;
    
    typedef pair<ll ,ll > pll;
    typedef pair<int ,int > pii;
    
    //priority_queue<int> q;//这是一个大根堆q
    //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
    #define fi first
    #define se second
    #define endl '
    '
    
    #define OKC ios::sync_with_stdio(false);cin.tie(0)
    #define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
    #define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
    //priority_queue<int ,vector<int>, greater<int> >que;
    
    const ll mos = 0x7FFFFFFF;  //2147483647
    const ll nmos = 0x80000000;  //-2147483648
    const int inf = 0x3f3f3f3f;
    const ll inff = 0x3f3f3f3f3f3f3f3f; //18
    
    template<typename T>
    inline T read(T&x){
        x=0;int f=0;char ch=getchar();
        while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
        while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
        return x=f?-x:x;
    }
    // #define _DEBUG;         //*//
    #ifdef _DEBUG
    freopen("input", "r", stdin);
    // freopen("output.txt", "w", stdout);
    #endif
    /*-----------------show time----------------*/
                    const int maxn = 1e5+9;
                    vector<int>g[maxn];
                    vector<int>v[maxn];
                    int cnt[maxn],col[maxn];
                    ll ans[maxn];
                    bool big[maxn];
                    int n;
                    int sz[maxn];
    
                    void dfs1(int v,int fa){
                        sz[v] = 1;
                        for(int i=0; i<g[v].size(); i++){
                            int u = g[v][i];
                            if(u!=fa){
                                dfs1(u,v);
                                sz[v]+=sz[u];
                            }
                        }
                    }
                    ll sum = 0,csz = 0;
                    // priority_queue<int>que;
                    void add(int v, int fa,int x){
    
                        cnt[col[v]] += x;
                        // if(v==1)cout<<"$$"<<cnt[col[v]]<<endl;
                        if(csz < cnt[col[v]]) sum = col[v] , csz = cnt[col[v]];
                        else if(csz==cnt[col[v]]) sum += col[v];
    
                        for(int i=0;i<g[v].size(); i++){
                            int u = g[v][i];
                            if(u!=fa && big[u]==0){
                                add(u,v,x);
                            }
                        }
                    }
    
                    void dfs(int v,int fa,int keep){
                        int mx = -1,bigChild = -1;
    
                        for(int i=0; i<g[v].size(); i++){
                            int u = g[v][i];
                            if(u!=fa && sz[u] > mx)
                                mx = sz[u], bigChild = u;
                        }
    
                        for(int i=0; i<g[v].size(); i++){
                            int u = g[v][i];
                            if(u!=fa && u!=bigChild){
                                dfs(u,v,0);////////////
                            }
                        }
    
                        if(bigChild!=-1){
                            dfs(bigChild,v,1),big[bigChild] = 1;/////////
                        }
    
                        add(v,fa,1);            ////////
                        ans[v] = sum;
                        /*
                        debug(v);
                        for(int i=1; i<=n; i++){
                            cout<<cnt[i]<<" ";
                        }
                        cout<<endl;
                        */
                        if(bigChild!=-1) big[bigChild] = 0;
                        if(keep==0){
                            add(v,fa,-1);
                            sum = 0;csz = 0;
                        }
                        /*
                        debug(v);
                        for(int i=1; i<=n; i++){
                            cout<<cnt[i]<<" ";
                        }
                        cout<<endl;
                        */
                    }
    
    int main(){
                    scanf("%d", &n);
                    for(int i=1; i<=n; i++)scanf("%d", &col[i]);
                    for(int i=1; i<n; i++){
                        int u,v;
                        scanf("%d%d", &u, &v);
                        g[u].pb(v);
                        g[v].pb(u);
                    }
                    dfs1(1,0);
                    dfs(1,0,1);
                    for(int i=1; i<=n; i++)printf("%lld " , ans[i]);
                    printf("
    ");
                    return 0;
    }
    EDU#2E
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  • 原文地址:https://www.cnblogs.com/ckxkexing/p/9425558.html
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