标签：+= min and clu app nes c++ mes shu
You are given a rooted tree with root in vertex 1. Each vertex is coloured in some colour.
Let‘s call colour c dominating in the subtree of vertex v if there are no other colours that appear in the subtree of vertex v more times than colour c. So it‘s possible that two or more colours will be dominating in the subtree of some vertex.
The subtree of vertex v is the vertex v and all other vertices that contains vertex v in each path to the root.
For each vertex v find the sum of all dominating colours in the subtree of vertex v.
The first line contains integer n (1 ≤ n ≤ 105) — the number of vertices in the tree.
The second line contains n integers ci (1 ≤ ci ≤ n), ci — the colour of the i-th vertex.
Each of the next n - 1 lines contains two integers xj, yj (1 ≤ xj, yj ≤ n) — the edge of the tree. The first vertex is the root of the tree.
Print n integers — the sums of dominating colours for each vertex.
4
1 2 3 4
1 2
2 3
2 4
10 9 3 4
15
1 2 3 1 2 3 3 1 1 3 2 2 1 2 3
1 2
1 3
1 4
1 14
1 15
2 5
2 6
2 7
3 8
3 9
3 10
4 11
4 12
4 13
6 5 4 3 2 3 3 1 1 3 2 2 1 2 3
思路：
对每个点建线段树在线段树上当前点的权值下标+1，每个点与父节点进行线段树合并，这样对树dfs一遍就可以求出所有的值了。
实现代码：
#include<bits/stdc++.h> using namespace std; #define ll long long #define mid ll m = (l + r) >> 1 const ll M = 1e5 + 10; struct node{ ll ans,sum; }tr[M*40]; struct node1{ ll to,next; }e[M<<1]; ll cnt,head[M],idx,ls[M*40],rs[M*40],root[M],a[M],ans[M]; void add(ll u,ll v){ e[++cnt].to = v;e[cnt].next = head[u];head[u] = cnt; } void pushup(ll rt){ if(tr[ls[rt]].sum > tr[rs[rt]].sum){ tr[rt].sum = tr[ls[rt]].sum; tr[rt].ans = tr[ls[rt]].ans; } else if(tr[ls[rt]].sum == tr[rs[rt]].sum){ tr[rt].sum = tr[ls[rt]].sum; tr[rt].ans = tr[rs[rt]].ans + tr[ls[rt]].ans; } else{ tr[rt].sum = tr[rs[rt]].sum; tr[rt].ans = tr[rs[rt]].ans; } } void update(ll p,ll c,ll l,ll r,ll &rt){ if(!rt) rt = ++idx; if(l == r){ tr[rt].sum += c; tr[rt].ans = l; return ; } mid; if(p <= m) update(p,c,l,m,ls[rt]); else update(p,c,m+1,r,rs[rt]); pushup(rt); } ll Merge(ll x,ll y,ll l,ll r){ if(!x) return y; if(!y) return x; if(l == r){ tr[x].sum += tr[y].sum; tr[x].ans = l; return x; } mid; ls[x] = Merge(ls[x],ls[y],l,m); rs[x] = Merge(rs[x],rs[y],m+1,r); pushup(x); return x; } void dfs(ll u,ll fa){ for(ll i = head[u];i;i=e[i].next){ ll v = e[i].to; if(v == fa) continue; dfs(v,u); Merge(root[u],root[v],1,M); } update(a[u],1,1,M,root[u]); ans[u] = tr[root[u]].ans; } int main() { ll n,u,v; scanf("%lld",&n); for(ll i = 1;i <= n;i ++){ scanf("%lld",&a[i]); root[i] = i; idx++; } for(ll i = 1;i < n;i ++){ scanf("%lld%lld",&u,&v); add(u,v); add(v,u); } dfs(1,0); for(ll i = 1;i <= n;i ++){ printf("%lld ",ans[i]); } return 0; }
codeforces 600E . Lomsat gelral (线段树合并）
标签：+= min and clu app nes c++ mes shu
原文地址：https://www.cnblogs.com/kls123/p/9972034.html
