cf600 E. Lomsat gelral

Posted 2021-09-07 Jozky86

cf600 E. Lomsat gelral

题意：

给出一个树，求出每个节点的子树中出现次数最多的颜色的编号和

题解：

树上启发式合并
树上启发式合并讲解
其实就是：递归轻儿子，消除影响，递归重儿子，不消除影响，递归轻儿子

void Dsu(int x,int fa,int opt){
	for(all Edge){
		if(to!=fa&&to!=BigSon[x])//暴力统计轻边的贡献
			dfs(to,x,0);
	}
	if(son[x])
		dfs(BigSon[x],x,1);//统计重儿子的贡献，不消除影响
	
	add(x);//暴力统计所有轻儿子的贡献
	ans[x]=NowAns;//更新答案
	if(!opt)delet(x);//删除刚开始统计轻边的影响
}

代码：

// Problem: E. Lomsat gelral
// Contest: Codeforces - Educational Codeforces Round 2
// URL: https://codeforces.com/contest/600/problem/E
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// Data:2021-08-18 23:54:38
// By Jozky

#include <bits/stdc++.h>
#include <unordered_map>
#define debug(a, b) printf("%s = %d\\n", a, b);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
clock_t startTime, endTime;
//Fe~Jozky
const ll INF_ll= 1e18;
const int INF_int= 0x3f3f3f3f;
void read(){};
template <typename _Tp, typename... _Tps> void read(_Tp& x, _Tps&... Ar)
{
    x= 0;
    char c= getchar();
    bool flag= 0;
    while (c < '0' || c > '9')
        flag|= (c == '-'), c= getchar();
    while (c >= '0' && c <= '9')
        x= (x << 3) + (x << 1) + (c ^ 48), c= getchar();
    if (flag)
        x= -x;
    read(Ar...);
}
template <typename T> inline void write(T x)
{
    if (x < 0) {
        x= ~(x - 1);
        putchar('-');
    }
    if (x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}
void rd_test()
{
#ifdef LOCAL
    startTime= clock();
    freopen("in.txt", "r", stdin);
#endif
}
void Time_test()
{
#ifdef LOCAL
    endTime= clock();
    printf("\\nRun Time:%lfs\\n", (double)(endTime - startTime) / CLOCKS_PER_SEC);
#endif
}
const int maxn= 3e5 + 9;
int n;
ll col[maxn];
int son[maxn], siz[maxn], cnt[maxn];
int Mx, Son;
ll sum, ans[maxn];
vector<int> vec[maxn];
void dfs1(int u, int fa)
{
    siz[u]= 1;
    for (auto it : vec[u]) {
        int v= it;
        if (v == fa)
            continue;
        dfs1(v, u);
        siz[u]+= siz[v];
        if (siz[v] > siz[son[u]])
            son[u]= v; //轻重链剖分
    }
}
void add(int x, int fa, int val)
{
    cnt[col[x]]+= val; //计算贡献

    if (cnt[col[x]] > Mx) {
        Mx= cnt[col[x]];
        sum= col[x];
    }
    else if (cnt[col[x]] == Mx) {
        sum+= col[x];
    }
    for (auto it : vec[x]) {
        int v= it;
        if (v == fa || v == Son)
            continue; //v是重儿子continue
        add(v, x, val); //继续统计
    }
}
void dfs2(int u, int fa, int opt)
{
    for (auto it : vec[u]) {
        int v= it;
        if (v != fa && v != son[u]) //说明是轻边
            dfs2(v, u, 0);
    }
    if (son[u]) {
        dfs2(son[u], u, 1); //统计重儿子的贡献
        Son= son[u];
    }
    add(u, fa, 1); //暴力统计所有轻儿子的贡献
    Son= 0;
    ans[u]= sum;
    if (!opt) {
        add(u, fa, -1); //消除贡献
        sum= 0;
        Mx= 0;
    }
}
int main()
{
    //rd_test();
    read(n);
    for (int i= 1; i <= n; i++)
        read(col[i]);
    for (int i= 1; i < n; i++) {
        int x, y;
        read(x, y);
        vec[x].push_back(y);
        vec[y].push_back(x);
    }
    dfs1(1, 0); //重链剖分
    dfs2(1, 0, 0);
    for (int i= 1; i <= n; i++) {
        printf("%lld ", ans[i]);
    }
    return 0;
    //Time_test();
}