CSU 1811 Tree Intersection(dfs序+莫队算法)

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1811: Tree Intersection

Time Limit: 5 Sec Memory Limit: 128 MB
Submit: 239 Solved: 108
[Submit][Status][Web Board]

Description

Bobo has a tree with n vertices numbered by 1,2,…,n and (n-1) edges. The i-th vertex has color c_i, and the i-th edge connects vertices a_i and b_i.

Let C(x,y) denotes the set of colors in subtree rooted at vertex x deleting edge (x,y).

Bobo would like to know R_i which is the size of intersection of C(a_i,b_i) and C(b_i,a_i) for all 1≤i≤(n-1). (i.e. |C(a_i,b_i)∩C(b_i,a_i)|)

Input

The input contains at most 15 sets. For each set:

The first line contains an integer n (2≤n≤10⁵).

The second line contains n integers c₁,c₂,…,c_n (1≤c_i≤n).

The i-th of the last (n-1) lines contains 2 integers a_i,b_i (1≤a_i,b_i≤n).

Output

For each set, (n-1) integers R₁,R₂,…,R_n-1.

Sample Input

41 2 2 11 22 33 451 1 2 1 21 32 33 54 5

Sample Output

题意：一棵树，每个节点有一种颜色，现在删去一条边，求这个树分成两部分，这两部分的共有颜色个数。

解题思路：首先我们可以知道，一颗子树的颜色种类数为tot，而它所特有的颜色种类为c，则分开后的两部分树的共有颜色为tot-p。

由于是点权的更新，而且关于子树的部分，所以可以先把它的dfs序弄出来。并且这道题并不需要更新节点权值，所以可以用莫队算法，而且莫队算法处理这种题很好写代码。

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<map>#include<algorithm>using namespace std;const int maxn = 100005;int n,sum[maxn],color[maxn],val[maxn];int block,cnt,L[maxn],R[maxn],res[maxn];vector<int> g[maxn];int Map[maxn];struct Edge{int u,v;}edge[maxn];struct Query{int l,r,id;bool operator < (const Query &rhs) const {if(l / block == rhs.l / block)              return r < rhs.r;          return l / block < rhs.l / block; }}q[maxn];void dfs(int u,int fa){L[u] = ++cnt;val[cnt] = color[u];for(int i = 0; i < g[u].size(); i++){if(g[u][i] == fa) continue;dfs(g[u][i],u);}R[u] = cnt;}void solve(){block = sqrt(cnt + 0.5);sort(q+1,q+n);int ans = 0,tot = 0,l = 1,r = 0;for(int i = 1; i < n; i++){if(q[i].l == q[i].r){if(sum[val[q[i].l]] == 1) res[q[i].id] = 0;else res[q[i].id] = 1;continue;}while(l < q[i].l){Map[val[l]]--;if(Map[val[l]] == sum[val[l]] - 1) ans--;if(Map[val[l]] == 0) tot--;l++;}while(l > q[i].l){l--;Map[val[l]]++;if(Map[val[l]] == sum[val[l]]) ans++;if(Map[val[l]] == 1) tot++;}while(r < q[i].r){r++;Map[val[r]]++;if(Map[val[r]] == sum[val[r]]) ans++;if(Map[val[r]] == 1) tot++;}while(r > q[i].r){Map[val[r]]--;if(Map[val[r]] == sum[val[r]] - 1) ans--;if(Map[val[r]] == 0) tot--;r--;}res[q[i].id] = tot - ans;}for(int i = 1; i < n; i++)printf("%d\n",res[i]);}int main(){int u,v;while(scanf("%d",&n)!=EOF){cnt = 0;memset(Map,0,sizeof(Map));memset(sum,0,sizeof(sum));for(int i = 1; i <= n; i++) g[i].clear();for(int i = 1; i <= n; i++){scanf("%d",&color[i]);sum[color[i]]++;}for(int i = 1; i < n; i++){scanf("%d %d",&u,&v);edge[i].u = u; edge[i].v = v;g[u].push_back(v);g[v].push_back(u);}dfs(1,-1);for(int i = 1; i < n; i++){u = edge[i].u;v = edge[i].v;if(L[u] < L[v]) swap(u,v);q[i].l = L[u];q[i].r = R[u];q[i].id = i;}solve();}return 0;}

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