Codeforces 600E (树上启发式合并) Posted on 2018-08-31 | In ACM , codeforces , 启发式合并 n个点的有根树,以1为根,每个点有一种颜色。我们称一种颜色占领了一个子树当且仅当没有其他颜色在这个子树中出现得比它多。求占领每个子树的所有颜色之和。(n≤105) 题解 树上启发式合并模板题 1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465666768697071727374757677#include <bits/stdc++.h>#define ll long longusing namespace std;const int maxn = 1e5+7;int n, col[maxn], st[maxn], ed[maxn], son[maxn], sz[maxn],cntt;vector<int>g[maxn];ll sum, ans[maxn], maxx , Son, cnt[maxn];void dfsxu(int x,int fa){ st[x]=++cntt; sz[x]=1; int mx=-1; for(int i=0;i<int(g[x].size());i++){ int v=g[x][i]; if(v==fa) continue; dfsxu(v, x); sz[x]+=sz[v]; if(sz[v] > mx){ mx=sz[v]; son[x]=v; } } ed[x]=cntt;}void solve(int x,int fa,int val){ cnt[col[x]]+=val; if(cnt[col[x]] > maxx) { maxx=cnt[col[x]]; sum = col[x]; } else if(cnt[col[x]] == maxx){ sum += col[x]; } for(int i=0;i<int(g[x].size());i++) { int v=g[x][i]; if(v==fa || v==Son) continue; solve(v, x , val); }}void dfs(int x,int fa, bool keep){ int mx=0; for(int i=0;i<int(g[x].size());i++) { int v=g[x][i]; if(v==fa||v==son[x]) continue; dfs(v, x, 0); } if(son[x]) dfs(son[x], x, 1), Son=son[x]; solve(x, fa, 1); Son=0; ans[x]=sum; if(!keep) solve(x,fa,-1), sum=maxx=0;}int main(){ int u, v; scanf("%d", &n); for(int i=1;i<=n;i++) scanf("%d", &col[i]); for(int i=1;i<=n-1;i++) { scanf("%d%d", &u, &v); g[u].push_back(v); g[v].push_back(u); } dfsxu(1, 0); dfs(1,0,0); for(int i=1;i<=n;i++) printf("%lld%c", ans[i], i==n?'\n':' '); return 0;}