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 LOCALstartTime= clock();freopen("in.txt", "r", stdin);
#endif
}
void Time_test()
{
#ifdef LOCALendTime= 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continueadd(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();
}