题目描述
You are given a rooted tree with root in vertex 1 1 1 . Each vertex is coloured in some colour.
Let's call colour c c c dominating in the subtree of vertex v v v if there are no other colours that appear in the subtree of vertex v v v more times than colour c c c . So it's possible that two or more colours will be dominating in the subtree of some vertex.
The subtree of vertex v v v is the vertex v v v and all other vertices that contains vertex v v v in each path to the root.
For each vertex v v v find the sum of all dominating colours in the subtree of vertex v v v .
输入输出格式
输入格式：
The first line contains integer n n n ( 1<=n<=105 1<=n<=10^{5} 1<=n<=105 ) — the number of vertices in the tree.
The second line contains n n n integers ci c_{i} ci ( 1<=ci<=n 1<=c_{i}<=n 1<=ci<=n ), ci c_{i} ci — the colour of the i i i -th vertex.
Each of the next n−1 n-1 n−1 lines contains two integers xj,yj x_{j},y_{j} xj,yj ( 1<=xj,yj<=n 1<=x_{j},y_{j}<=n 1<=xj,yj<=n ) — the edge of the tree. The first vertex is the root of the tree.
输出格式：
Print n n n integers — the sums of dominating colours for each vertex.
输入输出样例
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
题意：一棵树有n个结点，每个结点都是一种颜色，每个颜色有一个编号，求树中每个子树的最多的颜色编号的和。
题解：就是把每个点的颜色先插入每个点的对应的线段树里，然后dfs时把子树的线段树合并到父节点上，维护题意要求的值就可以了
代码如下：
#include<cstdio> #include<vector> #include<cstring> #include<iostream> #include<algorithm> #define lson tr[now].l #define rson tr[now].r #define int long long using namespace std; struct tree { int l,r,sum,val,ans; }tr[5000050]; int rt[100010],cl[100010],cnt,n,anss[100010]; vector<int> g[100010]; int push_up(int now) { if(tr[lson].sum>tr[rson].sum) { tr[now].sum=tr[lson].sum; tr[now].val=tr[lson].val; tr[now].ans=tr[lson].ans; } if(tr[rson].sum>tr[lson].sum) { tr[now].sum=tr[rson].sum; tr[now].val=tr[rson].val; tr[now].ans=tr[rson].ans; } if(tr[lson].sum==tr[rson].sum) { tr[now].sum=tr[lson].sum; tr[now].val=tr[lson].val; tr[now].ans=tr[lson].ans+tr[rson].ans; } } int update(int &now,int l,int r,int pos,int v) { if(!now) now=++cnt; if(l==r) { tr[now].val=l; tr[now].sum+=v; tr[now].ans=l; return 0; } int mid=(l+r)>>1; if(pos<=mid) update(lson,l,mid,pos,v); else update(rson,mid+1,r,pos,v); push_up(now); } int merge(int a,int b,int l,int r) { if(!a) return b; if(!b) return a; if(l==r) { tr[a].val=l; tr[a].sum+=tr[b].sum; tr[a].ans=l; return a; } int mid=(l+r)>>1; tr[a].l=merge(tr[a].l,tr[b].l,l,mid); tr[a].r=merge(tr[a].r,tr[b].r,mid+1,r); push_up(a); return a; } int dfs(int now,int f) { for(int i=0;i<g[now].size();i++) { if(g[now][i]==f) continue; dfs(g[now][i],now); merge(rt[now],rt[g[now][i]],1,100000); } update(rt[now],1,100000,cl[now],1); anss[now]=tr[rt[now]].ans; } int main() { scanf("%lld",&n); for(int i=1;i<=n;i++) { scanf("%lld",&cl[i]); rt[i]=i; cnt++; } int from,to; for(int i=1;i<n;i++) { scanf("%lld%lld",&from,&to); g[from].push_back(to); g[to].push_back(from); } dfs(1,0); for(int i=1;i<=n;i++) { printf("%lld ",anss[i]); } }