CSU 1663: Tree(树链剖分)
1663: Tree
Time Limit: 5 Sec Memory Limit: id=1663">Status pid=1663">Web
128 MB
Submit: 26 Solved: 11
[Submit][
Board
Description
CSU has a lot of trees. But there is a tree which is different from the others. This one is made of weighted edges and I have three kinds of operations on it:
1. C a b: Change the weight of edge a to b (1 ≤ b ≤ 100);
2. M a b c: Multiply the weights of those edges on the path from node a to node b by c (|c|≤10, c ≠ 0);
3. Q a b: Get the sum of weights from all the edges on the path from node a to node b.
Input
There are multiple test cases.
The first line will contain a positive integer T (T ≤ 10) meaning the number of test cases.
Each test case will have an integer N (1 ≤ N ≤ 50,000) indicating the number of nodes marked from 1 to N.
Then N-1 lines followed. Each lines contains three integers a, b and c (1 ≤ a, b, c ≤ N, a ≠ b, 1 ≤ c ≤ 100) indicating the there is an edge connecting node a and node b with weight c. The edges are marked from 1 to N-1 in the order of appearance.
Then some operations followed, with one operation per line. The format is as shown in the problem description. A single letter ‘E’ indicates the end of the case. There are no more than 50,000 operations per test case.
Output
For each ‘Q’ operation, output an integer meaning the sum. The final result will never exceed 32-bit signed integer.
Sample Input
2
4
1 2 14
1 3 20
3 4 1
C 3 11
M 1 4 3
M 3 4 2
M 1 3 2
Q 2 4
M 1 4 -1
M 3 4 -2
Q 2 4
E
3
1 2 4
1 3 2
Q 2 3
C 1 7
M 2 3 2
M 1 3 5
Q 2 3
E
Sample Output
200
26
6
34
#include<stdio.h>
#include<string.h>
#define LL long long
const int N = 50005; int head[N<<1],to[N<<1],next1[N<<1],tot;
int deep[N],fath[N],son[N],num[N];
int top[N],p[N],pos; void init(){
pos=tot=0;
memset(head,-1,sizeof(head));
}
void addEdge(const int& u, const int& v){
to[tot] = v, next1[tot] = head[u], head[u] = tot++;
}
void addUndirEdge(const int& u, const int& v){
addEdge(u, v), addEdge(v, u);
} void dfs1(int u,int pre,int d){
fath[u]=pre;
deep[u]=d;
son[u]=-1;
num[u]=1;
for(int i=head[u]; i!=-1; i=next1[i]){
int v=to[i];
if(v==fath[u])continue;
dfs1(v,u,d+1);
num[u]+=num[v];
if(son[u]==-1||num[v]>num[son[u]])
son[u]=v;
}
}
void getpos(int u,int root){
top[u]=root;
p[u]=pos++;
if(son[u]==-1)
return ;
getpos(son[u],root);
for(int i=head[u]; i!=-1; i=next1[i]){
int v=to[i];
if(v==son[u]||v==fath[u])
continue;
getpos(v,v);
}
}
struct TREE{
LL milt,sum;
}root[N*3];
LL cost[N]; void pushUp(int k){
root[k].sum=root[k<<1].sum+root[k<<1|1].sum;
}
void pushDow(int k){
if(root[k].milt!=1)
{
root[k<<1].sum*=root[k].milt;
root[k<<1].milt*=root[k].milt; root[k<<1|1].sum*=root[k].milt;
root[k<<1|1].milt*=root[k].milt;
root[k].milt=1;
}
}
void build(int l, int r, int k){
root[k].milt=1;
if(l==r){
root[k].sum=cost[l]; return ;
}
int mid=(l+r)>>1;
build(l,mid,k<<1);
build(mid+1,r,k<<1|1);
pushUp(k);
}
void update_C(int l, int r, int k, const int& id, LL c){
if(l==r){
root[k].sum=c; return ;
}
int mid=(l+r)>>1;
pushDow(k);
if(id<=mid)
update_C(l,mid,k<<1,id,c);
else
update_C(mid+1,r,k<<1|1,id,c);
pushUp(k);
}
void updata_M(int l,int r,int k,int L,int R,int M){
if(L<=l&&r<=R){
root[k].milt*=M; root[k].sum*=M;
return ;
}
pushDow(k);
int mid=(l+r)>>1;
if(L<=mid)
updata_M(l,mid,k<<1,L,R,M);
if(mid<R)
updata_M(mid+1,r,k<<1|1,L,R,M);
pushUp(k);
}
LL query(int l, int r, int k, const int& L, const int& R){
if(L<=l&&r<=R){
return root[k].sum;
}
pushDow(k);
int mid=(l+r)>>1;
LL sum=0;
if(L<=mid)
sum+=query(l,mid,k<<1,L,R);
if(mid<R)
sum+=query(mid+1,r,k<<1|1,L,R);
return sum;
}
void swp(int &u,int &v){
int tt=u; u=v; v=tt;
}
LL solve(int u,int v,int flag,LL M){
int fu=top[u], fv=top[v];
LL sum=0;
while(fu!=fv){
if(deep[fu]<deep[fv]){
swp(fu,fv); swp(u,v);
}
if(flag==0)
updata_M(1,pos,1,p[fu],p[u],M);
else
sum+=query(1,pos,1,p[fu],p[u]);
u=fath[fu]; fu=top[u];
}
if(u==v)return sum;
if(deep[u]>deep[v])
swp(u,v);
if(flag==0)
updata_M(1,pos,1,p[son[u]],p[v],M);
else
sum+=query(1,pos,1,p[son[u]],p[v]);//一不小心p[son[u]]写成了p[u]让我WA了好几次(求边权用p[son[u]],求点权用p[u])
return sum;
} struct EDG{
int u,v;
LL c;
}edg[N]; int main()
{
int n,m,a,b,T;
char op[10];
scanf("%d",&T);
while(T--){
scanf("%d",&n);
init();
for(int i=1; i<n; i++){
scanf("%d%d%lld",&edg[i].u,&edg[i].v,&edg[i].c);
addUndirEdge(edg[i].u, edg[i].v);
}
dfs1(1,1,1);
getpos(1,1);
for(int i=1; i<n; i++){
if(deep[edg[i].u]>deep[edg[i].v])
swp(edg[i].u, edg[i].v);
cost[p[edg[i].v]]=edg[i].c;
}
pos=n;
build(1,pos,1); while(1){
scanf("%s",op);
if(op[0]=='E')
break;
scanf("%d%d",&a,&b);
if(op[0]=='C')
update_C(1,pos,1,p[edg[a].v],b);
else if(op[0]=='M'){
LL M;
scanf("%lld",&M);
solve(a,b,0,M);
}
else
printf("%lld\n",solve(a,b,1,1));
}
} } /**************************************************************
Problem: 1663
User: aking2015
Language: C++
Result: Accepted
Time:2696 ms
Memory:9044 kb
****************************************************************/
CSU 1663: Tree(树链剖分)的更多相关文章
- Hdu 5274 Dylans loves tree (树链剖分模板)
Hdu 5274 Dylans loves tree (树链剖分模板) 题目传送门 #include <queue> #include <cmath> #include < ...
- POJ3237 Tree 树链剖分 边权
POJ3237 Tree 树链剖分 边权 传送门:http://poj.org/problem?id=3237 题意: n个点的,n-1条边 修改单边边权 将a->b的边权取反 查询a-> ...
- Query on a tree——树链剖分整理
树链剖分整理 树链剖分就是把树拆成一系列链,然后用数据结构对链进行维护. 通常的剖分方法是轻重链剖分,所谓轻重链就是对于节点u的所有子结点v,size[v]最大的v与u的边是重边,其它边是轻边,其中s ...
- 【BZOJ-4353】Play with tree 树链剖分
4353: Play with tree Time Limit: 20 Sec Memory Limit: 256 MBSubmit: 31 Solved: 19[Submit][Status][ ...
- SPOJ Query on a tree 树链剖分 水题
You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, ...
- poj 3237 Tree 树链剖分
题目链接:http://poj.org/problem?id=3237 You are given a tree with N nodes. The tree’s nodes are numbered ...
- Codeforces Round #200 (Div. 1) D Water Tree 树链剖分 or dfs序
Water Tree 给出一棵树,有三种操作: 1 x:把以x为子树的节点全部置为1 2 x:把x以及他的所有祖先全部置为0 3 x:询问节点x的值 分析: 昨晚看完题,马上想到直接树链剖分,在记录时 ...
- poj 3237 Tree 树链剖分+线段树
Description You are given a tree with N nodes. The tree’s nodes are numbered 1 through N and its edg ...
- Aizu 2450 Do use segment tree 树链剖分+线段树
Do use segment tree Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://www.bnuoj.com/v3/problem_show ...
随机推荐
- openfire源码解读之将cache和session对象移入redis以提升性能
原文:http://blog.csdn.net/jinzhencs/article/details/50522322 前言: 目前我们的openfire服务器只能支撑单机2W 集群4W.(估测在线用户 ...
- 【spring Boot】Spring中@Controller和@RestController之间的区别
spring Boot入手的第一天,看到例子中的@RestController ............. 相同点:都是用来表示Spring某个类的是否可以接收HTTP请求 不同点:@Controll ...
- Javascript 内置值、typeof运算符、true/false判断
一.内置值 true false null undefined NaN Infinity 二.typeof运算结果 ...
- appium python学习记录
这是网上找到的测试用例 from appium import webdriver desired_caps = {} desired_caps['platformName'] = 'Android' ...
- CentOS7.4 x64环境Percona-Server-5.6安装
CentOS7.4 x64环境Percona-Server-5.6安装 下载MySQL $ cd /usr/local/src/ $ wget https://www.percona.com/down ...
- windows 用wireshark抓本机的包
原文: http://bijian1013.iteye.com/blog/2299856 1.也可以用另外一个工具: RawCap 当然也不是说windows下就别想抓到本地回路的包了,肯定有别的方法 ...
- Elasticsearch教程(一),全程直播(小白级别)
ElasticSearch是一个基于Lucene的搜索服务器.它提供了一个分布式多用户能力的全文搜索引擎,基于RESTful web接口.Elasticsearch是用Java开发的,并作为Apach ...
- EffectiveJava(20)使用子类型化优化标签类
标签类:其中有许多样板代码,包括枚举声明,标签域和条件语句 如果要给它添加风格,除了有权限修改源码之外,你还得给每个条件语句都添加一个条件,否则就会在运行时失败 标签类过于冗长,容易出错,并且效率低下 ...
- Win7如何开启Messenger服务
1 如图所示,在WIN7系统中没有找到Messenger这个服务.因为Messenger在Windows 7/Server 2008 R2里都去掉了 2 而在XP系统中,开启了Messenger服 ...
- linux配置jdk失败
在linux下配置jdk时,/etc/profile下的配置内容是对的,可是输入java -version却发现配置没有成功,这一般都是jdk的安装文件夹权限没有提升的原因,仅仅需用chmod -R ...