标签：handle div == rate app from upd strong def
You are given a tree with N nodes. The tree nodes are numbered from 1 to N and have colors C1, C2,
. . . , CN initially. You have to handle M instructions on the tree of the following forms:
• 0 u c: Change the color of node u to c.
• 1 u v: Output the maximum number of times a color appeared on the unique path from node u
to node v.
Input
The first line of input contains T (1 ≤ T ≤ 10) which is the number of test cases. The first line of each
test case contains two integers N and M (1 ≤ N, M ≤ 105
). Next line contains N space separated
integers C1, C2, . . . , CN (1 ≤ Ci ≤ 10) denoting the initial colors of the nodes. Each of the next N − 1
lines contain two integers a and b (1 ≤ a, b ≤ N and a ?= b) meaning that there is an edge between
node a and node b. Each of the next M lines contains an instruction of one of the two forms described
above. For all the instructions: 1 ≤ u, v ≤ N and 1 ≤ c ≤ 10.
Output
For each of the second type instruction output the answer in one line.
Sample Input
2
5 6
3 2 1 2 3
1 2
2 3
2 4
1 5
1 3 5
0 1 1
0 2 1
1 3 5
0 2 4
1 2 4
2 1
5 6
1 2
1 2 2
Sample Output
2
3
1
1
题意：给你一颗树，树上的每个节点都有颜色，颜色编号在1-10之间，给出两种操作
操作一：将树上x点的颜色搞成val
操作二：求点u到点v路径上最多的颜色有几个
题解：颜色很少，所以不用树上莫队，可以建10棵线段树，如果一个点的颜色为ai，则在ai编号的线段树上该点为1，其他编号的树上该点为0
然后每次跑路径的时候跑十遍，记下路径和最大的一种颜色输出，就是答案了,emmm,真暴力啊QAQ
因为涉及到树上路径，所以树链剖分即可了
代码如下：
#include<cstdio> #include<vector> #include<cstring> #include<iostream> #include<algorithm> #define lson root<<1 #define rson root<<1|1 using namespace std; struct seg_tree { struct node { int l,r,sum; } tr[400040]; void push_up(int root) { tr[root].sum=tr[lson].sum+tr[rson].sum; } void build(int root,int l,int r) { if(l==r) { tr[root].l=l; tr[root].r=r; return ; } tr[root].l=l; tr[root].r=r; int mid=(l+r)>>1; build(lson,l,mid); build(rson,mid+1,r); } void update(int root,int pos,int val) { if(tr[root].l==pos&&tr[root].r==pos) { tr[root].sum=val; return ; } int mid=(tr[root].l+tr[root].r)>>1; if(pos<=mid) { update(lson,pos,val); } else { update(rson,pos,val); } push_up(root); } int query(int root,int l,int r) { if(tr[root].l==l&&tr[root].r==r) { return tr[root].sum; } int mid=(tr[root].l+tr[root].r)>>1; if(mid<l) { return query(rson,l,r); } else { if(mid>=r) { return query(lson,l,r); } else { return query(lson,l,mid)+query(rson,mid+1,r); } } } void dfs(int root,int l,int r) { if(l==r) { return ; } int mid=(l+r)>>1; dfs(lson,l,mid); dfs(rson,mid+1,r); } } tree[11]; vector<int> g[100010]; int fa[100010],deep[100010],size[100010],son[100010],id[100010],c[100010],w[100010],top[100010],cnt; void dfs1(int now,int f,int dep) { fa[now]=f; deep[now]=dep; size[now]=1; int maxson=-1; for(int i=0; i<g[now].size(); i++) { if(g[now][i]==f) { continue; } dfs1(g[now][i],now,dep+1); size[now]+=size[g[now][i]]; if(size[g[now][i]]>maxson) { maxson=size[g[now][i]]; son[now]=g[now][i]; } } } void dfs2(int now,int topf) { id[now]=++cnt; w[cnt]=c[now]; top[now]=topf; if(!son[now]) { return ; } dfs2(son[now],topf); for(int i=0; i<g[now].size(); i++) { if(fa[now]==g[now][i]||g[now][i]==son[now]) { continue; } dfs2(g[now][i],g[now][i]); } } void point_update(int u,int val) { for(int i=1; i<=10; i++) { if(val==i) { tree[i].update(1,id[u],1); } else { tree[i].update(1,id[u],0); } } } int path_query(int u,int v,int val) { int ans=0; while(top[u]!=top[v]) { if(deep[top[u]]<deep[top[v]]) { swap(u,v); } ans+=tree[val].query(1,id[top[u]],id[u]); u=fa[top[u]]; } if(deep[u]>deep[v]) { swap(u,v); } ans+=tree[val].query(1,id[u],id[v]); return ans; } int main() { int t; scanf("%d",&t); while(t--) { int n,m; memset(tree,0,sizeof(tree)); memset(fa,0,sizeof(fa)); memset(son,0,sizeof(son)); memset(deep,0,sizeof(deep)); memset(size,0,sizeof(size)); memset(id,0,sizeof(id)); memset(c,0,sizeof(c)); memset(w,0,sizeof(w)); memset(top,0,sizeof(top)); cnt=0; for(int i=1;i<=100000;i++) { g[i].clear(); } scanf("%d%d",&n,&m); for(int i=1; i<=n; i++) { scanf("%d",&c[i]); } for(int i=1; i<n; i++) { int from,to; scanf("%d%d",&from,&to); g[from].push_back(to); g[to].push_back(from); } dfs1(1,0,1); dfs2(1,1); for(int i=1; i<=10; i++) { tree[i].build(1,1,n); } for(int i=1; i<=n; i++) { tree[w[i]].update(1,i,1); } for(int i=1; i<=m; i++) { int kd,l,r; scanf("%d%d%d",&kd,&l,&r); if(kd==1) { int ans=0; for(int j=1; j<=10; j++) { ans=max(ans,path_query(l,r,j)); } printf("%d\n",ans); } else { if(kd==0) { point_update(l,r); } } } } }
UVA - 12424 Answering Queries on a Tree(十棵线段树的树链剖分)
标签：handle div == rate app from upd strong def
原文地址：https://www.cnblogs.com/stxy-ferryman/p/9026117.html
