2002: [Hnoi2010]Bounce 弹飞绵羊

https://www.lydsy.com/JudgeOnline/problem.php?id=2002

分析:

  绵羊在弹飞的路径中相当于一棵树,这棵树需要更改形态,删一条边,加一条边,所以LCT维护一下。

代码:

 #include<bits/stdc++.h>
using namespace std;
typedef long long LL; inline int read() {
int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
for (;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
} const int N = ;
int ch[N][],fa[N],rev[N],val[N],siz[N],sk[N],a[N],Top,n; void pushup(int x) {
siz[x] = siz[ch[x][]] + siz[ch[x][]] + ;
}
void pushdown(int x) {
if (rev[x]) {
rev[ch[x][]] ^= ; rev[ch[x][]] ^= ;
swap(ch[x][], ch[x][]);
rev[x] ^= ;
}
}
bool isroot(int x) {
return ch[fa[x]][] != x && ch[fa[x]][] != x;
}
int son(int x) {
return x == ch[fa[x]][];
}
void rotate(int x) {
int y = fa[x],z = fa[y],b = son(x),c = son(y),a = ch[x][!b];
if (!isroot(y)) ch[z][c] = x;fa[x] = z;
ch[x][!b] = y;fa[y] = x;
ch[y][b] = a;if (a) fa[a] = y;
pushup(y);pushup(x);
}
void splay(int x) {
sk[Top = ] = x;
for (int i=x; !isroot(i); i=fa[i]) sk[++Top] = fa[i];
while (Top) pushdown(sk[Top--]);
while (!isroot(x)) {
int y = fa[x];
if (isroot(y)) rotate(x);
else {
if (son(x) == son(y)) rotate(y), rotate(x);
else rotate(x), rotate(x);
}
}
}
void access(int x) {
for (int last=; x; last=x, x=fa[x]) {
splay(x); ch[x][] = last; pushup(x);
}
}
void makeroot(int x) {
access(x); splay(x); rev[x] ^= ;
}
int find(int x) {
access(x); splay(x);
while (ch[x][]) x = ch[x][];
return x;
}
void link(int x,int y) {
makeroot(x);
fa[x] = y;
}
void cut(int x,int y) {
makeroot(x); access(y); splay(y);
if (fa[x] == y && !ch[x][]) fa[x] = ch[y][] = ;
} void query() {
makeroot(n + );
int p = read() + ;
access(p);
splay(p);
printf("%d\n",siz[p] - );
}
void change() {
int p = read() + ,x = read(),t = p+x > n+ ? n+: p+x; // --- 判断是否大于n+1,不判luogu上80
cut(p,a[p]);
a[p] = t;
link(p,a[p]);
} int main() {
n = read();
for (int i=; i<=n; ++i) {
a[i] = i + read();
a[i] = a[i] > n+ ? n+ : a[i];
}
for (int i=; i<=n; ++i) link(i,a[i]);
int m = read();
while (m--) {
int opt = read();
if (opt == ) query();
else change();
}
return ;
}

2002: [Hnoi2010]Bounce 弹飞绵羊的更多相关文章

  1. BZOJ 2002: [Hnoi2010]Bounce 弹飞绵羊

    2002: [Hnoi2010]Bounce 弹飞绵羊 Time Limit: 10 Sec  Memory Limit: 259 MBSubmit: 9071  Solved: 4652[Submi ...

  2. BZOJ 2002: [Hnoi2010]Bounce 弹飞绵羊 分块

    2002: [Hnoi2010]Bounce 弹飞绵羊 Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://www.lydsy.com/JudgeOn ...

  3. BZOJ 2002: [Hnoi2010]Bounce 弹飞绵羊 LCT

    2002: [Hnoi2010]Bounce 弹飞绵羊 Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://www.lydsy.com/JudgeOn ...

  4. bzoj 2002: [Hnoi2010]Bounce 弹飞绵羊 動態樹

    2002: [Hnoi2010]Bounce 弹飞绵羊 Time Limit: 10 Sec  Memory Limit: 259 MBSubmit: 4055  Solved: 2172[Submi ...

  5. bzoj 2002 : [Hnoi2010]Bounce 弹飞绵羊 (LCT)

    链接:https://www.lydsy.com/JudgeOnline/problem.php?id=2002 题面: 2002: [Hnoi2010]Bounce 弹飞绵羊 Time Limit: ...

  6. BZOJ 2002: [Hnoi2010]Bounce 弹飞绵羊 (动态树LCT)

    2002: [Hnoi2010]Bounce 弹飞绵羊 Time Limit: 10 Sec  Memory Limit: 259 MBSubmit: 2843  Solved: 1519[Submi ...

  7. 【BZOJ】2002: [Hnoi2010]Bounce 弹飞绵羊

    2002: [Hnoi2010]Bounce 弹飞绵羊 Time Limit: 10 Sec  Memory Limit: 259 MBSubmit: 14802  Solved: 7507[Subm ...

  8. BZOJ 2002: [Hnoi2010]Bounce 弹飞绵羊 【分块】

    任意门:https://www.lydsy.com/JudgeOnline/problem.php?id=2002 2002: [Hnoi2010]Bounce 弹飞绵羊 Time Limit: 10 ...

  9. 2002: [Hnoi2010]Bounce 弹飞绵羊(分块)

    2002: [Hnoi2010]Bounce 弹飞绵羊 时间限制: 10 Sec  内存限制: 259 MB 题目描述 某天,Lostmonkey发明了一种超级弹力装置,为了在他的绵羊朋友面前显摆,他 ...

随机推荐

  1. BZOJ1718:[USACO]Redundant Paths 分离的路径(双连通分量)

    Description In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numb ...

  2. Jupyter notebook 的一个问题

    Traceback (most recent call last): File , in get value = obj._trait_values[self.name] KeyError: 'all ...

  3. 生理周期,POJ(1006)

    题目链接:http://poj.org/problem?id=1006 解题报告: 1.枚举天数的时候可以根据前面的结果直接跳过一些错误的答案. ///三个周期是23,28,33, #include ...

  4. c#类的练习

    类部分练习题 - dijiaxing1234的博客 - CSDN博客  https://blog.csdn.net/dijiaxing1234/article/details/81230811 真好啊

  5. 【洛谷P2341】受欢迎的牛

    受欢迎的牛 题目描述 一些可以当明星的牛,一定会构成一个强连通分量,我们可以先缩点,最后统计一下出度为零的强连通分量大小即可, 若出度为零的强连通分量个数大于1,则输出0 #include<io ...

  6. EJB结合struts2创建项目、发布jboss服务器和访问、父类(BaseDaoImpl)的封装

    一.环境搭建: 1.准备jboss服务器,将对应数据库的xml配置好放到jboss的发布目录下. <?xml version="1.0" encoding="UTF ...

  7. centOS 7 更改root密码

    Linux忘记密码怎么办,不用重装系统,进入emergency mode 更改root密码即可. 首先重启系统,按下 向下 按钮, 定位在第一个,摁 e ,进行编辑 找到 ro , 把ro改为 rw ...

  8. Angularjs 数据过滤

    <!DOCTYPE html><html><head><meta http-equiv="Content-Type" content=&q ...

  9. c语言描述的直接插入排序法

    #include<stdio.h> #include<stdlib.h> #define SIZE 6 typedef int Type; //直接插入排序法 void Ins ...

  10. VSTO Project 客户端自动发送邮件

    利用office vsto功能,抓取我们选择的任务,根据配置节,邮件发送内容,最终根据任务名称,任务开始结束时间,任务资源名称,发送邮件给任务资源. 这是我的VSTO界面. 配置我们发送邮件的服务器地 ...