题目1:BZOJ 2049 洞穴勘测

 #include <bits/stdc++.h>
#define L(x) c[x][0]
#define R(x) c[x][1] using namespace std; const int oo = 0x3f3f3f3f; struct SplayTree{
static const int N = + ; int top, st[N];
int fa[N], c[N][], sum[N], mx[N], val[N];
bool rev[N]; bool isroot(int x){
return L(fa[x]) != x && R(fa[x]) != x;
} void pushdown(int x){
int l = L(x), r = R(x); if(rev[x]){
rev[x] ^= ; rev[l] ^= ; rev[r] ^= ;
swap(L(x), R(x));
}
} void pushup(int x){
int l = L(x), r = R(x); sum[x] = sum[l] + sum[r] + val[x];
mx[x] = max(val[x], max(mx[l], mx[r]));
} void rotate(int x){
int y = fa[x], z = fa[y], l, r; l = (L(y) == x) ^ ; r = l ^ ;
if(!isroot(y))
c[z][(L(z) == y) ^ ] = x;
fa[x] = z; fa[y] = x; fa[c[x][r]] = y;
c[y][l] = c[x][r]; c[x][r] = y;
} void splay(int x){
top = ;
st[++ top] = x;
for(int i = x; !isroot(i); i = fa[i])
st[++ top] = fa[i];
while(top) pushdown(st[top --]);
while(!isroot(x)){
int y = fa[x], z = fa[y]; if(!isroot(y)){
if(L(y) == x ^ L(z) == y) rotate(x);
else rotate(y);
}
rotate(x);
}
}
}Splay; struct LinkCutTree{ void Access(int x){
int t = ; while(x){
Splay.splay(x);
Splay.R(x) = t;
t = x; x = Splay.fa[x];
}
} void makeroot(int x){
Access(x); Splay.splay(x); Splay.rev[x] ^= ;
} void link(int x, int y){
makeroot(x); Splay.fa[x] = y; Splay.splay(x);
} void cut(int x, int y){
makeroot(x); Access(y); Splay.splay(y);
Splay.L(y) = Splay.fa[x] = ;
} int findroot(int x){
Access(x); Splay.splay(x);
int y = x;
while(Splay.L(y)) y = Splay.L(y); return y;
}
}lct; int n, m, x, y;
char str[]; int main(){
scanf("%d%d", &n, &m);
for(int i = ; i <= m; ++ i){
scanf("%s%d%d", str, &x, &y); if(str[] == 'C') lct.link(x, y);
else if(str[] == 'D') lct.cut(x, y);
else{
if(lct.findroot(x) == lct.findroot(y)) printf("Yes\n");
else printf("No\n");
}
} return ;
}

BZOJ 2049

题目2: BZOJ 1180 OTOCI

 #include <bits/stdc++.h>
#define L(x) c[x][0]
#define R(x) c[x][1] using namespace std; typedef long long ll;
const int N = + ;
const int oo = 0x3f3f3f3f; struct SplayTree{
int fa[N], c[N][];
ll val[N], mx[N], sum[N];
bool rev[N];
int top, st[N]; inline bool isroot(int x){
return L(fa[x]) != x && R(fa[x]) != x;
} void pushdown(int x){
int l = L(x), r = R(x); if(rev[x]){
rev[x] ^= ; rev[l] ^= ; rev[r] ^= ;
swap(L(x), R(x));
}
} void pushup(int x){
int l = L(x), r = R(x); sum[x] = sum[l] + sum[r] + val[x];
mx[x] = max(val[x], max(mx[l], mx[r]));
} void rotate(int x){
int y = fa[x], z = fa[y], l, r; l = (L(y) == x) ^ ; r = l ^ ;
if(!isroot(y)){
c[z][(L(z) == y) ^ ] = x;
}
fa[x] = z; fa[y] = x; fa[c[x][r]] = y;
c[y][l] = c[x][r]; c[x][r] = y;
pushup(y); pushup(x);
} void splay(int x){
top = ;
st[++ top] = x; for(int i = x; !isroot(i); i = fa[i])
st[++ top] = fa[i];
while(top) pushdown(st[top --]);
while(!isroot(x)){
int y = fa[x], z = fa[y]; if(!isroot(y)){
if(L(y) == x ^ L(z) == y) rotate(x);
else rotate(y);
}
rotate(x);
}
}
}Splay; struct LinkCutTree{
void Access(int x){
int t = ; while(x){
Splay.splay(x);
Splay.R(x) = t;
t = x;
Splay.pushup(x);
x = Splay.fa[x];
}
} void makeroot(int x){
Access(x); Splay.splay(x); Splay.rev[x] ^= ;
} void link(int x, int y){
makeroot(x); Splay.fa[x] = y; Splay.splay(x);
} void cut(int x, int y){
makeroot(x); Access(y); Splay.splay(y);
Splay.L(y) = Splay.fa[x] = ;
} int findroot(int x){
Access(x); Splay.splay(x); int y = x; while(Splay.L(y)){
y = Splay.L(y);
} return y;
}
}lct; int n, m, x, y;
char str[]; int main(){
scanf("%d", &n);
for(int i = ; i <= n; ++ i){
scanf("%lld", &Splay.val[i]);
Splay.sum[i] = Splay.val[i];
}
scanf("%d", &m);
for(int i = ; i <= m; ++ i){
scanf("%s%d%d", str, &x, &y); if(str[] == 'b'){
if(lct.findroot(x) == lct.findroot(y))
puts("no");
else{
puts("yes"); lct.link(x, y);
}
}
else if(str[] == 'e'){
if(lct.findroot(x) != lct.findroot(y)){
puts("impossible");
}
else{
lct.makeroot(x); lct.Access(y); Splay.splay(y);
printf("%lld\n", Splay.sum[y]);
}
}
else{
lct.makeroot(x); Splay.val[x] = y; Splay.pushup(x);
}
} return ;
}

BZOJ 1180

题目3:BZOJ 3282 TREE

 #include <bits/stdc++.h>
#define L(x) c[x][0]
#define R(x) c[x][1] using namespace std; const int N = + ; struct SplayTree{
int fa[N], c[N][], val[N], sum[N];
bool rev[N];
int top, st[N]; inline bool isroot(int x){
return L(fa[x]) != x && R(fa[x]) != x;
} void pushup(int x){
int l = L(x), r = R(x); sum[x] = sum[l] ^ sum[r] ^ val[x];
}
void pushdown(int x){
int l = L(x), r = R(x); if(rev[x]){
rev[x] ^= ; rev[l] ^= ; rev[r] ^= ;
swap(L(x), R(x));
}
}
void rotate(int x){
int y = fa[x], z = fa[y], l, r; l = (L(y) == x) ^ ; r = l ^ ; if(!isroot(y)){
c[z][(L(z) == y) ^ ] = x;
}
fa[y] = x; fa[x] = z; fa[c[x][r]] = y;
c[y][l] = c[x][r]; c[x][r] = y;
pushup(y); pushup(x);
} void splay(int x){
top = ;
st[++ top] = x;
for(int i = x; !isroot(i); i = fa[i])
st[++ top] = fa[i]; while(top) pushdown(st[top --]);
while(!isroot(x)){
int y = fa[x], z = fa[y]; if(!isroot(y)){
if(L(y) == x ^ L(z) == y) rotate(x);
else rotate(y);
}
rotate(x);
}
}
}Splay; struct LinkCutTree{
void Access(int x){
int t = ; while(x){
Splay.splay(x);
Splay.R(x) = t;
t = x;
Splay.pushup(x);
x = Splay.fa[x];
}
} void makeroot(int x){
Access(x); Splay.splay(x); Splay.rev[x] ^= ;
} int findroot(int x){
Access(x); Splay.splay(x); int y = x; while(Splay.L(y)) y = Splay.L(y); return y;
} void link(int x, int y){
makeroot(x); Splay.fa[x] = y; Splay.splay(x);
} void cut(int x, int y){
makeroot(x); Access(y); Splay.splay(y);
Splay.L(y) = Splay.fa[x] = ;
}
}lct; int n, m;
int x, y, z; int main(){
scanf("%d%d", &n, &m);
for(int i = ; i <= n; ++ i){
scanf("%d", &Splay.val[i]);
Splay.sum[i] = Splay.val[i];
}
for(int i = ; i <= m; ++ i){
scanf("%d%d%d", &z, &x, &y); switch(z){
case :{
lct.makeroot(x); lct.Access(y); Splay.splay(y);
printf("%d\n", Splay.sum[y]);
break;
}
case :{
if(lct.findroot(x) != lct.findroot(y)){
lct.link(x, y);
}
break;
}
case :{
if(lct.findroot(x) == lct.findroot(y)){
lct.cut(x, y);
}
break;
}
case :{
lct.Access(x); Splay.splay(x); Splay.val[x] = y;
Splay.pushup(x);
break;
}
}
} return ;
}

BZOJ 3282

题目4: HDU 4010 Query On The Trees

 #include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <iostream>
#define L(x) c[x][0]
#define R(x) c[x][1] using namespace std; const int N = + ;
const int oo = 0x3f3f3f3f; int n, m, tot;
int first[N], nxt[N<<];
int u[N<<], v[N<<];
int que[N<<]; struct SplayTree{
int fa[N], c[N][], val[N], mx[N], add[N];
bool rev[N];
int top, st[N]; inline bool isroot(int x){
return L(fa[x]) != x && R(fa[x]) != x;
} void pushup(int x){
int l = L(x), r = R(x); mx[x] = max(mx[l], mx[r]);
mx[x] = max(mx[x], val[x]);
} void pushdown(int x){
int l = L(x), r = R(x); if(rev[x]){
rev[x] ^= ; rev[l] ^= ; rev[r] ^= ;
swap(L(x), R(x));
}
if(add[x]){
if(l){
add[l] += add[x]; mx[l] += add[x]; val[l] += add[x];
}
if(r){
add[r] += add[x]; mx[r] += add[x]; val[r] += add[x];
}
add[x] = ;
}
} void rotate(int x){
int y = fa[x], z = fa[y], l, r; l = (L(y) == x) ^ ; r = l ^ ;
if(!isroot(y)){
c[z][L(z) == y ^ ] = x;
}
fa[x] = z; fa[y] = x; fa[c[x][r]] = y;
c[y][l] = c[x][r]; c[x][r] = y;
pushup(y); pushup(x);
} void splay(int x){
top = ;
st[++ top] = x;
for(int i = x; !isroot(i); i = fa[i])
st[++ top] = fa[i];
while(top) pushdown(st[top --]);
while(!isroot(x)){
int y = fa[x], z = fa[y]; if(!isroot(y)){
if(L(y) == x ^ L(z) == y) rotate(x);
else rotate(y);
}
rotate(x);
}
}
}Splay; struct LinkCutTree{
void Access(int x){
int t = ; while(x){
Splay.splay(x);
Splay.R(x) = t;
t = x;
Splay.pushup(x);
x = Splay.fa[x];
}
}
void Add(int x, int y, int w){
makeroot(x); Access(y); Splay.splay(y);
Splay.add[y] += w; Splay.val[y] += w;
Splay.mx[y] += w;
}
int findroot(int x){
Access(x); Splay.splay(x); int y = x; while(Splay.L(y)){
y = Splay.L(y);
} return y;
} void makeroot(int x){
Access(x); Splay.splay(x); Splay.rev[x] ^= ;
} void cut(int x, int y){
makeroot(x); Access(y); Splay.splay(y);
Splay.L(y) = Splay.fa[Splay.L(y)] = ; Splay.pushup(y);
} void link(int x, int y){
makeroot(x); Splay.fa[x] = y; Splay.splay(x);
}
}lct; void insert(int s, int t){
++ tot;
u[tot] = s; v[tot] = t;
nxt[tot] = first[s];
first[s] = tot;
} void bfs(){
int head, tail;
head = tail = ;
que[head] = ;
Splay.fa[] = ; while(head <= tail){
int x = que[head]; for(int i = first[x]; i; i = nxt[i]){
if(v[i] != Splay.fa[x]){
Splay.fa[v[i]] = x;
que[++ tail] = v[i];
}
}
++ head;
}
} int main(){
int flag, x, y, z; while(~scanf("%d", &n)){
tot = ;
for(int i = ; i <= n; ++ i){
Splay.fa[i] = Splay.val[i] = Splay.add[i] = ;
Splay.rev[i] = false;
Splay.mx[i] = -oo;
Splay.c[i][] = Splay.c[i][] = ;
first[i] = ;
}
for(int i = ; i < n; ++ i){
scanf("%d%d", &x, &y);
insert(x, y); insert(y, x);
}
for(int i = ; i <= n; ++ i){
scanf("%d", &x);
Splay.val[i] = Splay.mx[i] = x;
}
bfs(); scanf("%d", &m);
for(int i = ; i <= m; ++ i){
scanf("%d", &flag);
switch(flag){
case :{
scanf("%d%d", &x, &y);
if(lct.findroot(x) == lct.findroot(y))
puts("-1");
else
lct.link(x, y);
break;
}
case :{
scanf("%d%d", &x, &y);
if(lct.findroot(x) != lct.findroot(y) || x == y)
puts("-1");
else lct.cut(x, y);
break;
}
case :{
scanf("%d%d%d", &z, &x, &y);
if(lct.findroot(x) != lct.findroot(y))
puts("-1");
else
lct.Add(x, y, z);
break;
}
case :{
scanf("%d%d", &x, &y);
if(lct.findroot(x) != lct.findroot(y))
puts("-1");
else{
lct.makeroot(x); lct.Access(y); Splay.splay(y);
printf("%d\n", Splay.mx[y]);
}
break;
}
}
}
puts("");
} return ;
}

HDU 4010

Link-Cut-Tree题目泛做(为了对应自己的课件)的更多相关文章

  1. K-D Tree题目泛做(CXJ第二轮)

    题目1: BZOJ 2716 题目大意:给出N个二维平面上的点,M个操作,分为插入一个新点和询问到一个点最近点的Manhatan距离是多少. 算法讨论: K-D Tree 裸题,有插入操作. #inc ...

  2. Codeforces Round #339 (Div. 2) A. Link/Cut Tree 水题

    A. Link/Cut Tree 题目连接: http://www.codeforces.com/contest/614/problem/A Description Programmer Rostis ...

  3. 洛谷P3690 [模板] Link Cut Tree [LCT]

    题目传送门 Link Cut Tree 题目背景 动态树 题目描述 给定n个点以及每个点的权值,要你处理接下来的m个操作.操作有4种.操作从0到3编号.点从1到n编号. 0:后接两个整数(x,y),代 ...

  4. LCT总结——概念篇+洛谷P3690[模板]Link Cut Tree(动态树)(LCT,Splay)

    为了优化体验(其实是强迫症),蒟蒻把总结拆成了两篇,方便不同学习阶段的Dalao们切换. LCT总结--应用篇戳这里 概念.性质简述 首先介绍一下链剖分的概念(感谢laofu的讲课) 链剖分,是指一类 ...

  5. Link Cut Tree学习笔记

    从这里开始 动态树问题和Link Cut Tree 一些定义 access操作 换根操作 link和cut操作 时间复杂度证明 Link Cut Tree维护链上信息 Link Cut Tree维护子 ...

  6. Link Cut Tree 总结

    Link-Cut-Tree Tags:数据结构 ##更好阅读体验:https://www.zybuluo.com/xzyxzy/note/1027479 一.概述 \(LCT\),动态树的一种,又可以 ...

  7. link cut tree 入门

    鉴于最近写bzoj还有51nod都出现写不动的现象,决定学习一波厉害的算法/数据结构. link cut tree:研究popoqqq那个神ppt. bzoj1036:维护access操作就可以了. ...

  8. [CodeForces - 614A] A - Link/Cut Tree

    A - Link/Cut Tree Programmer Rostislav got seriously interested in the Link/Cut Tree data structure, ...

  9. 【刷题】洛谷 P3690 【模板】Link Cut Tree (动态树)

    题目背景 动态树 题目描述 给定n个点以及每个点的权值,要你处理接下来的m个操作.操作有4种.操作从0到3编号.点从1到n编号. 0:后接两个整数(x,y),代表询问从x到y的路径上的点的权值的xor ...

  10. 脑洞大开加偏执人格——可持久化treap版的Link Cut Tree

    一直没有点动态树这个科技树,因为听说只能用Splay,用Treap的话多一个log.有一天脑洞大开,想到也许Treap也能从底向上Split.仔细思考了一下,发现翻转标记不好写,再仔细思考了一下,发现 ...

随机推荐

  1. Bootstrap新手常见问题

    题记 bootstrap这个开源的UI库确实很方便,用了两日,觉得不错,但也有些问题比较头疼! 主题 1.怎么使用?怎么定制?下面是一个范例,修改了navbar的颜色,重新设置了select控件的默认 ...

  2. poj3579 二分搜索+二分查找

    Median Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5468   Accepted: 1762 Descriptio ...

  3. C++类中的静态成员变量与静态成员函数的使用

    代码: #include <iostream> #include <string> #include <cstdio> using namespace std; c ...

  4. hdu5348 MZL's endless loop(欧拉回路)

    转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud MZL's endless loop Time Limit: 3000/1500 ...

  5. paip.输入法编程----一级汉字1000个

    paip.输入法编程----一级汉字1000个.txt 作者Attilax ,  EMAIL:1466519819@qq.com  来源:attilax的专栏 地址:http://blog.csdn. ...

  6. Aphache VFS

    http://blog.csdn.net/hemingwang0902/article/details/4733911 http://jackyrong.iteye.com/blog/1330946 ...

  7. Oracle11g R2学习系列 之四Maven+Struts+Spring实验

    今天试一下Java调用Oracle来看一下.会不会也如昨天实验的一样坑呢?由于我对于Java也接触的不多,所以不打算直接使用该收提供的实验文档,而是自己利用Maven+Struts+Spring来自己 ...

  8. Song of Pi

    def main(): pi = ' # 预先给出需要比较的值 t = int(raw_input()) for _ in xrange(t): song = raw_input().strip(). ...

  9. unable to find valid certification path to requested target

    Error : javax.net.ssl.SSLHandshakeException: sun.security.validator.ValidatorException: PKIX path bu ...

  10. 使用ARM和VMSS创建自动扩展的web集群

    在很多的商业场景中,用户的访问,峰值时间都是很难预测的,尤其是做一些市场推广活动和促销的时候,到底部署什么规模的web集群合适,这一直是个问题,部署过量会造成高成本和资源不必要的浪费,部署过少,如果到 ...