[bzoj2002][Hnoi2010]Bounce 弹飞绵羊 分块 Link-Cut-Tree

2002: [Hnoi2010]Bounce 弹飞绵羊

Time Limit: 10 Sec  Memory Limit: 259 MB

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Description

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，他邀请小绵羊一起玩个游戏。游戏一开始，Lostmonkey在地上沿着一条直线摆上n个装置，每个装置设定初始弹力系数ki，当绵羊达到第i个装置时，它会往后弹ki步，达到第i+ki个装置，若不存在第i+ki个装置，则绵羊被弹飞。绵羊想知道当它从第i个装置起步时，被弹几次后会被弹飞。为了使得游戏更有趣，Lostmonkey可以修改某个弹力装置的弹力系数，任何时候弹力系数均为正整数。

Input

第一行包含一个整数n，表示地上有n个装置，装置的编号从0到n-1,接下来一行有n个正整数，依次为那n个装置的初始弹力系数。第三行有一个正整数m，接下来m行每行至少有两个数i、j，若i=1，你要输出从j出发被弹几次后被弹飞，若i=2则还会再输入一个正整数k，表示第j个弹力装置的系数被修改成k。对于20%的数据n,m<=10000，对于100%的数据n<=200000,m<=100000

Output

对于每个i=1的情况，你都要输出一个需要的步数，占一行。

Sample Input

4

1 2 1 1

3

1 1

2 1 1

1 1

Sample Output

2

3

HINT

Source

本来是学LCT的，结果看到这道题就写了个分块（因为写不来LCT）。
每次复杂度为根号n
维护在每个区间的步数和到下一个区间的数

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
const int N = 200000 + 5;
int n,m,k
,l[1005],belong
,block,cnt,ans;
int next
,ste
;
int main(){
scanf("%d", &n); block = sqrt(n); cnt = n%block ? n/block+1 : block;
for( int i = 1; i <= n; i++ ) scanf("%d", &k[i]);
for( int i = 1; i <= cnt; i++ ) l[i] = (i-1)*block+1;
for( int i = 1; i <= n; i++ ) belong[i] = (i-1)/block+1;
for( int i = n; i >= 1; i-- ){
if( i + k[i] > n ) ste[i] = 1;
else if( belong[i] == belong[i+k[i]] ){
next[i] = next[i+k[i]]; ste[i] = ste[i+k[i]]+1;
} else next[i] = i + k[i], ste[i] = 1;
}
scanf("%d", &m);
for( int i = 1,flag,ch,cto; i <= m; i++ ){
scanf("%d", &flag);
if( flag == 1 ){
scanf("%d", &ch); ch++; ans = 0;
for( int j = ch; j; j = next[j] ) ans += ste[j];
printf("%d\n", ans);
} else {
scanf("%d%d", &ch, &cto); k[++ch] = cto;
for( int j = ch; j >= l[belong[ch]]; j-- ){
if( belong[j] == belong[j+k[j]] ) next[j] = next[j+k[j]], ste[j] = ste[j+k[j]]+1;
else next[j] = j + k[j], ste[j] = 1;
}
}
}
return 0;
}

再也不相信文化课了（期末炸翻天）
7.12
LCT版
每次操作相当于在树上剪去或增加一棵树

#include<iostream>
#include<cstdio>
using namespace std;
const int N = 200005;
int n,m,fa
,sta
,siz
,c
[2],next
;
bool rev
; char ch[10];
bool isroot( int x ){
return c[fa[x]][1]!=x && c[fa[x]][0]!=x;
}
void updata( int k ){
siz[k] = siz[c[k][0]] + siz[c[k][1]] + 1;
}
void pushdown( int k ){
if( rev[k] ){
swap(c[k][1],c[k][0]);
rev[c[k][1]] ^= 1; rev[c[k][0]] ^= 1;
rev[k] = 0;
}
}
void rotate( int x ){
int y = fa[x], z = fa[y], l, r;
l = c[y][0] == x ? 0 : 1; r = l^1;
if( !isroot(y) ){
if( c[z][0]==y ) c[z][0] = x;
else c[z][1] = x;
}
fa[x] = z; fa[y] = x; fa[c[x][r]] = y;
c[y][l] = c[x][r]; c[x][r] = y;
updata(x); updata(y);
}
void splay( int x ){
int top = 0; sta[++top] = x;
for( int i = x; !isroot(i); i = fa[i] ) sta[++top] = fa[i];
for( int i = top; i; i-- ) pushdown(sta[i]);
while( !isroot(x) ){
int y = fa[x], z = fa[y];
if( !isroot(y) ){
if( (c[z][0]==y)^(c[y][0]==x) ) rotate(x);
else rotate(y);
}
rotate(x);
}
}
void access( int x ){
int t = 0;
while(x){
splay(x);
c[x][1] = t;
t = x; x = fa[t];
}
}
void mtr( int x ){
access(x); splay(x); rev[x] ^= 1;
}
void link( int x, int y ){
mtr(x); fa[x] = y;
}
void cut( int x, int y ){
mtr(x); access(y); splay(y); c[y][0] = fa[x] = 0;
}
int find( int x ){
int y = x; access(x); splay(x);
while( c[y][0] ) y = c[y][0];
return y;
}
int main(){
scanf("%d", &n );
for( int i = 1,x; i <= n; i++ ){
scanf("%d", &x);
next[i] = fa[i] = x+i>n ? n+1 : x+i; siz[x] = 1;
} siz[n+1] = 1;
scanf("%d", &m);
for( int i = 1,f,x,y; i <= m; i++ ){
scanf("%d", &f );
if( f == 1 ){
scanf("%d", &x); x++;
mtr(n+1); access(x);
splay(x); printf("%d\n", siz[c[x][0]]);
} else{
scanf("%d%d", &x, &y); x++;
int to = min(n+1,x+y);
cut( x, next[x] ); link( x, to ); next[x] = to;
}
}
return 0;
}