SPOJ 10628. Count on a tree （树上第k大，LCA+主席树）

10628. Count on a tree

Problem code: COT

 

You are given a tree with N nodes.The tree nodes are numbered from 1 to N.Each node has an integer weight.

We will ask you to perform the following operation:

  • u v k : ask for the kth minimum weight on the path from node u to node v

 

Input

In the first line there are two integers N and M.(N,M<=100000)

In the second line there are N integers.The ith integer denotes the weight of the ith node.

In the next N-1 lines,each line contains two integers u v,which describes an edge (u,v).

In the next M lines,each line contains three integers u v k,which means an operation asking for the kth minimum weight on the path from node u to node v.

Output

For each operation,print its result.

Example

Input:
8 5
8 5
105 2 9 3 8 5 7 7
1 2        
1 3
1 4
3 5
3 6
3 7
4 8
2 5 1
2 5 2
2 5 3
2 5 4
7 8 2 
Output:
2
8
9
105
7 

 

 

在树上建立主席树。

然后求LCA

  1 /* ***********************************************
  2 Author        :kuangbin
  3 Created Time  :2013-9-5 10:31:57
  4 File Name     :F:\2013ACM练习\专题学习\主席树\SPOJ_COT.cpp
  5 ************************************************ */
  6 
  7 #include <stdio.h>
  8 #include <string.h>
  9 #include <iostream>
 10 #include <algorithm>
 11 #include <vector>
 12 #include <queue>
 13 #include <set>
 14 #include <map>
 15 #include <string>
 16 #include <math.h>
 17 #include <stdlib.h>
 18 #include <time.h>
 19 using namespace std;
 20 
 21 //主席树部分 *****************8
 22 const int MAXN = 200010;
 23 const int M = MAXN * 40;
 24 int n,q,m,TOT;
 25 int a[MAXN], t[MAXN];
 26 int T[M], lson[M], rson[M], c[M];
 27 
 28 void Init_hash()
 29 {
 30     for(int i = 1; i <= n;i++)
 31         t[i] = a[i];
 32     sort(t+1,t+1+n);
 33     m = unique(t+1,t+n+1)-t-1;
 34 }
 35 int build(int l,int r)
 36 {
 37     int root = TOT++;
 38     c[root] = 0;
 39     if(l != r)
 40     {
 41         int mid = (l+r)>>1;
 42         lson[root] = build(l,mid);
 43         rson[root] = build(mid+1,r);
 44     }
 45     return root;
 46 }
 47 int hash(int x)
 48 {
 49     return lower_bound(t+1,t+1+m,x) - t;
 50 }
 51 int update(int root,int pos,int val)
 52 {
 53     int newroot = TOT++, tmp = newroot;
 54     c[newroot] = c[root] + val;
 55     int l = 1, r = m;
 56     while( l < r)
 57     {
 58         int mid = (l+r)>>1;
 59         if(pos <= mid)
 60         {
 61             lson[newroot] = TOT++; rson[newroot] = rson[root];
 62             newroot = lson[newroot]; root = lson[root];
 63             r = mid;
 64         }
 65         else
 66         {
 67             rson[newroot] = TOT++; lson[newroot] = lson[root];
 68             newroot = rson[newroot]; root = rson[root];
 69             l = mid+1;
 70         }
 71         c[newroot] = c[root] + val;
 72     }
 73     return tmp;
 74 }
 75 int query(int left_root,int right_root,int LCA,int k)
 76 {
 77     int lca_root = T[LCA];
 78     int pos = hash(a[LCA]);
 79     int l = 1, r = m;
 80     while(l < r)
 81     {
 82         int mid = (l+r)>>1;
 83         int tmp = c[lson[left_root]] + c[lson[right_root]] - 2*c[lson[lca_root]] + (pos >= l && pos <= mid);
 84         if(tmp >= k)
 85         {
 86             left_root = lson[left_root];
 87             right_root = lson[right_root];
 88             lca_root = lson[lca_root];
 89             r = mid;
 90         }
 91         else
 92         {
 93             k -= tmp;
 94             left_root = rson[left_root];
 95             right_root = rson[right_root];
 96             lca_root = rson[lca_root];
 97             l = mid + 1;
 98         }
 99     }
100     return l;
101 }
102 
103 //LCA部分
104 int rmq[2*MAXN];//rmq数组，就是欧拉序列对应的深度序列
105 struct ST
106 {
107     int mm[2*MAXN];
108     int dp[2*MAXN][20];//最小值对应的下标
109     void init(int n)
110     {
111         mm[0] = -1;
112         for(int i = 1;i <= n;i++)
113         {
114             mm[i] = ((i&(i-1)) == 0)?mm[i-1]+1:mm[i-1];
115             dp[i][0] = i;
116         }
117         for(int j = 1; j <= mm[n];j++)
118             for(int i = 1; i + (1<<j) - 1 <= n; i++)
119                 dp[i][j] = rmq[dp[i][j-1]] < rmq[dp[i+(1<<(j-1))][j-1]]?dp[i][j-1]:dp[i+(1<<(j-1))][j-1];
120     }
121     int query(int a,int b)//查询[a,b]之间最小值的下标
122     {
123         if(a > b)swap(a,b);
124         int k = mm[b-a+1];
125         return rmq[dp[a][k]] <= rmq[dp[b-(1<<k)+1][k]]?dp[a][k]:dp[b-(1<<k)+1][k];
126     }
127 };
128 //边的结构体定义
129 struct Edge
130 {
131     int to,next;
132 };
133 Edge edge[MAXN*2];
134 int tot,head[MAXN];
135 
136 int F[MAXN*2];//欧拉序列，就是dfs遍历的顺序，长度为2*n-1,下标从1开始
137 int P[MAXN];//P[i]表示点i在F中第一次出现的位置
138 int cnt;
139 
140 ST st;
141 void init()
142 {
143     tot = 0;
144     memset(head,-1,sizeof(head));
145 }
146 void addedge(int u,int v)//加边，无向边需要加两次
147 {
148     edge[tot].to = v;
149     edge[tot].next = head[u];
150     head[u] = tot++;
151 }
152 void dfs(int u,int pre,int dep)
153 {
154     F[++cnt] = u;
155     rmq[cnt] = dep;
156     P[u] = cnt;
157     for(int i = head[u];i != -1;i = edge[i].next)
158     {
159         int v = edge[i].to;
160         if(v == pre)continue;
161         dfs(v,u,dep+1);
162         F[++cnt] = u;
163         rmq[cnt] = dep;
164     }
165 }
166 void LCA_init(int root,int node_num)//查询LCA前的初始化
167 {
168     cnt = 0;
169     dfs(root,root,0);
170     st.init(2*node_num-1);
171 }
172 int query_lca(int u,int v)//查询u,v的lca编号
173 {
174     return F[st.query(P[u],P[v])];
175 }
176 
177 void dfs_build(int u,int pre)
178 {
179     int pos = hash(a[u]);
180     T[u] = update(T[pre],pos,1);
181     for(int i = head[u]; i != -1;i = edge[i].next)
182     {
183         int v = edge[i].to;
184         if(v == pre)continue;
185         dfs_build(v,u);
186     }
187 }
188 int main()
189 {
190     //freopen("in.txt","r",stdin);
191     //freopen("out.txt","w",stdout);
192     while(scanf("%d%d",&n,&q) == 2)
193     {
194         for(int i = 1;i <= n;i++)
195             scanf("%d",&a[i]);
196         Init_hash();
197         init();
198         TOT = 0;
199         int u,v;
200         for(int i = 1;i < n;i++)
201         {
202             scanf("%d%d",&u,&v);
203             addedge(u,v);
204             addedge(v,u);
205         }
206         LCA_init(1,n);
207         T[n+1] = build(1,m);
208         dfs_build(1,n+1);
209         int k;
210         while(q--)
211         {
212             scanf("%d%d%d",&u,&v,&k);
213             printf("%d\n",t[query(T[u],T[v],query_lca(u,v),k)]);
214         }
215         return 0;
216     }
217     return 0;
218 }

 

 

 

 

 

 

posted on 2013-09-07 07:42  kuangbin  阅读(2080)  评论(0)  编辑  收藏  举报

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