SPOJ 375. Query on a tree （树链剖分）

题目链接：

http://www.spoj.com/problems/QTREE/

 

375. Query on a tree

Problem code: QTREE

 

You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1.

We will ask you to perfrom some instructions of the following form:

  • CHANGE i ti : change the cost of the i-th edge to ti
    or
  • QUERY a b : ask for the maximum edge cost on the path from node a to node b

Input

The first line of input contains an integer t, the number of test cases (t <= 20). t test cases follow.

For each test case:

  • In the first line there is an integer N (N <= 10000),
  • In the next N-1 lines, the i-th line describes the i-th edge: a line with three integers a b c denotes an edge between a, b of cost c (c <= 1000000),
  • The next lines contain instructions "CHANGE i ti" or "QUERY a b",
  • The end of each test case is signified by the string "DONE".

There is one blank line between successive tests.

Output

For each "QUERY" operation, write one integer representing its result.

Example

Input:
1

3
1 2 1
2 3 2
QUERY 1 2
CHANGE 1 3
QUERY 1 2
DONE

Output:
1
3

 

树链剖分入门题：

 

 

  1 /* **********************************************
  2 Author      : kuangbin
  3 Created Time: 2013/8/11 22:00:02
  4 File Name   : F:\2013ACM练习\专题学习\数链剖分\SPOJ_QTREE.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 using namespace std;
 19 
 20 const int MAXN = 10010;
 21 struct Edge
 22 {
 23     int to,next;
 24 }edge[MAXN*2];
 25 int head[MAXN],tot;
 26 int top[MAXN];//top[v]表示v所在的重链的顶端节点
 27 int fa[MAXN]; //父亲节点
 28 int deep[MAXN];//深度
 29 int num[MAXN];//num[v]表示以v为根的子树的节点数
 30 int p[MAXN];//p[v]表示v与其父亲节点的连边在线段树中的位置
 31 int fp[MAXN];//和p数组相反
 32 int son[MAXN];//重儿子
 33 int pos;
 34 void init()
 35 {
 36     tot = 0;
 37     memset(head,-1,sizeof(head));
 38     pos = 0;
 39     memset(son,-1,sizeof(son));
 40 }
 41 void addedge(int u,int v)
 42 {
 43     edge[tot].to = v;edge[tot].next = head[u];head[u] = tot++;
 44 }
 45 void dfs1(int u,int pre,int d) //第一遍dfs求出fa,deep,num,son
 46 {
 47     deep[u] = d;
 48     fa[u] = pre;
 49     num[u] = 1;
 50     for(int i = head[u];i != -1; i = edge[i].next)
 51     {
 52         int v = edge[i].to;
 53         if(v != pre)
 54         {
 55             dfs1(v,u,d+1);
 56             num[u] += num[v];
 57             if(son[u] == -1 || num[v] > num[son[u]])
 58                 son[u] = v;
 59         }
 60     }
 61 }
 62 void getpos(int u,int sp) //第二遍dfs求出top和p
 63 {
 64     top[u] = sp;
 65     if(son[u] != -1)
 66     {
 67         p[u] = pos++;
 68         fp[p[u]] = u;
 69         getpos(son[u],sp);
 70     }
 71     else
 72     {
 73         p[u] = pos++;
 74         fp[p[u]] = u;
 75         return;
 76     }
 77     for(int i = head[u] ; i != -1; i = edge[i].next)
 78     {
 79         int v = edge[i].to;
 80         if(v != son[u] && v != fa[u])
 81             getpos(v,v);
 82     }
 83 }
 84 
 85 //线段树
 86 struct Node
 87 {
 88     int l,r;
 89     int Max;
 90 }segTree[MAXN*3];
 91 void build(int i,int l,int r)
 92 {
 93     segTree[i].l = l;
 94     segTree[i].r = r;
 95     segTree[i].Max = 0;
 96     if(l == r)return;
 97     int mid = (l+r)/2;
 98     build(i<<1,l,mid);
 99     build((i<<1)|1,mid+1,r);
100 }
101 void push_up(int i)
102 {
103     segTree[i].Max = max(segTree[i<<1].Max,segTree[(i<<1)|1].Max);
104 }
105 void update(int i,int k,int val) // 更新线段树的第k个值为val
106 {
107     if(segTree[i].l == k && segTree[i].r == k)
108     {
109         segTree[i].Max = val;
110         return;
111     }
112     int mid = (segTree[i].l + segTree[i].r)/2;
113     if(k <= mid)update(i<<1,k,val);
114     else update((i<<1)|1,k,val);
115     push_up(i);
116 }
117 int query(int i,int l,int r)  //查询线段树中[l,r] 的最大值
118 {
119     if(segTree[i].l == l && segTree[i].r == r)
120         return segTree[i].Max;
121     int mid = (segTree[i].l + segTree[i].r)/2;
122     if(r <= mid)return query(i<<1,l,r);
123     else if(l > mid)return query((i<<1)|1,l,r);
124     else return max(query(i<<1,l,mid),query((i<<1)|1,mid+1,r));
125 }
126 int find(int u,int v)//查询u->v边的最大值
127 {
128     int f1 = top[u], f2 = top[v];
129     int tmp = 0;
130     while(f1 != f2)
131     {
132         if(deep[f1] < deep[f2])
133         {
134             swap(f1,f2);
135             swap(u,v);
136         }
137         tmp = max(tmp,query(1,p[f1],p[u]));
138         u = fa[f1]; f1 = top[u];
139     }
140     if(u == v)return tmp;
141     if(deep[u] > deep[v]) swap(u,v);
142     return max(tmp,query(1,p[son[u]],p[v]));
143 }
144 int e[MAXN][3];
145 int main()
146 {
147     //freopen("in.txt","r",stdin);
148     //freopen("out.txt","w",stdout);
149     int T;
150     int n;
151     scanf("%d",&T);
152     while(T--)
153     {
154         init();
155         scanf("%d",&n);
156         for(int i = 0;i < n-1;i++)
157         {
158             scanf("%d%d%d",&e[i][0],&e[i][1],&e[i][2]);
159             addedge(e[i][0],e[i][1]);
160             addedge(e[i][1],e[i][0]);
161         }
162         dfs1(1,0,0);
163         getpos(1,1);
164         build(1,0,pos-1);
165         for(int i = 0;i < n-1; i++)
166         {
167             if(deep[e[i][0]] > deep[e[i][1]])
168                 swap(e[i][0],e[i][1]);
169             update(1,p[e[i][1]],e[i][2]);
170         }
171         char op[10];
172         int u,v;
173         while(scanf("%s",op) == 1)
174         {
175             if(op[0] == 'D')break;
176             scanf("%d%d",&u,&v);
177             if(op[0] == 'Q')
178                 printf("%d\n",find(u,v));
179             else update(1,p[e[u-1][1]],v);
180         }
181     }
182     return 0;
183 }

 

 

posted on 2013-08-12 19:45  kuangbin  阅读(3600)  评论(2)  编辑  收藏  举报

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