spoj 375 QTREE - Query on a tree 树链剖分 LCT 动态树
2015-08-21 10:45
573 查看
QTREE - Query on a tree
no tags 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
分析:
对每个点建立一个结点,值为0
对每个边建立一个结点,值为边权。
然后用LCT维护即可
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<vector> using namespace std; #define maxn 500007 #define inf 1000000000 #define ll int struct Node{ Node *fa,*ch[2]; bool rev,root; int val; ll minv; }; Node pool[maxn]; Node *nil,*tree[maxn]; int cnt = 0; void init(){ cnt = 1; nil = tree[0] = pool; nil->ch[0] = nil->ch[1] = nil; nil->val = 0; nil->minv = 0; } Node *newnode(int val,Node *f){ pool[cnt].fa = f; pool[cnt].ch[0]=pool[cnt].ch[1]=nil; pool[cnt].rev = false; pool[cnt].root = true; pool[cnt].val = val; pool[cnt].minv = val; return &pool[cnt++]; } //左右子树反转******真正把结点变为根 void update_rev(Node *x){ if(x == nil) return ; x->rev = !x->rev; swap(x->ch[0],x->ch[1]); } //splay向上更新信息****** void update(Node *x){ if(x == nil) return ; x->minv = x->val; Node*y = x->ch[0]; if(y->minv > x->minv) x->minv = y->minv; y = x->ch[1]; if(y->minv > x->minv) x->minv = y->minv; } //splay下推信息****** void pushdown(Node *x){ if(x->rev != false){ update_rev(x->ch[0]); update_rev(x->ch[1]); x->rev = false; } } //splay在root-->x的路径下推信息****** void push(Node *x){ if(!x->root) push(x->fa); pushdown(x); } //将结点x旋转至splay中父亲的位置****** void rotate(Node *x){ Node *f = x->fa, *ff = f->fa; int t = (f->ch[1] == x); if(f->root) x->root = true, f->root = false; else ff->ch[ff->ch[1] == f] = x; x->fa = ff; f->ch[t] = x->ch[t^1]; x->ch[t^1]->fa = f; x->ch[t^1] = f; f->fa = x; update(f); } //将结点x旋转至x所在splay的根位置****** void splay(Node *x){ push(x); Node *f, *ff; while(!x->root){ f = x->fa,ff = f->fa; if(!f->root) if((ff->ch[1]==f)&&(f->ch[1] == x)) rotate(f); else rotate(x); rotate(x); } update(x); } //将x到树根的路径并成一条path****** Node *access(Node *x){ Node *y = nil,*z; while(x != nil){ splay(x); x->ch[1]->root = true; (x->ch[1] = y)->root = false; update(x); y = x; x = x->fa; } return y; } //将结点x变成树根****** void be_root(Node *x){ access(x); splay(x); update_rev(x); } //将x连接到结点f上****** void link(Node *x, Node *f){ be_root(x); x->fa = f; } Node * find(Node *root){ if(root->ch[0] == nil) return root; return find(root->ch[0]); } Node*road[maxn]; int main(){ int n,q,t; Node*x,*y,*z; scanf("%d",&t); char word[20]; while(t--){ scanf("%d",&n); init(); int u,v,t; for(int i = 1;i <= n; i++) tree[i] = newnode(0,nil); for(int i = 1;i < n ;i++){ scanf("%d%d%d",&u,&v,&t); road[i] = x = newnode(t,nil); link(tree[u],x); link(tree[v],x); } while(1){ scanf("%s",word); if(word[0] == 'D') break; scanf("%d%d",&u,&v); if(word[0] == 'Q'){ be_root(tree[u]); y = access(tree[v]); printf("%d\n",y->minv); } else { splay(road[u]); road[u]->val = v; update(road[u]); } } } return 0; } /* 3 100 1 1 2 1 2 3 2 0 2 1 2 3 0 3 1 2 5 0 2 */
相关文章推荐
- UIDeviceOrientation和UIInterfaceOrientation中left、right的含义
- Pop Sequence (25)--栈模拟经典
- The type java.util.Map$Entry cannot be resolved. It is indirectly referenced from required .class
- [LeetCode#52]N-Queens II
- Android开源资料大集合_架构&UI
- apue源码运行配置
- iOS UIViewController 和 nib 相关的3个方法
- Java之String,StringBuffer,StringBuilder类
- JAVA-QUEUE类图
- NGUI小细节 生成与点击事件
- UVA 540 Team Queue
- iOS中使用storyboard实现页面跳转,Segue详解及简单的数据传递
- 【Druid】 阿里巴巴推出的国产数据库连接池com.alibaba.druid.pool.DruidDataSource
- POJ 2299 Ultra-QuickSort 归并排序
- UIButton设置圆角
- query builder 在线生成sql
- The request sent by the client was syntactically incorrect.
- Android项目使用Ant打包,自动生成build.xml
- Win10预览版10525有哪些新功能 Win10 Build 10525更新内容汇总
- Error:No suitable device found: no device found for connection "System eth0"