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。
BZOJ 2002([Hnoi2010]Bounce 弹飞绵羊-LCT)
2002: [Hnoi2010]Bounce 弹飞绵羊
Time Limit: 10 Sec Memory Limit: 259 MBSubmit: 5212 Solved: 2774
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Description
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
1 2 1 1
3
1 1
2 1 1
1 1
Sample Output
2
3
3
HINT
Source
Splay 启发式合并
第一题LCT
参考《QTREE解法的一些研究》
和kuangbin的代码
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (200000+10)
#define MAXM (100000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int n,m;
class Splay
{
public:
int father[MAXN],siz[MAXN];
int ch[MAXN][2];
bool root[MAXN];
void mem(int n)
{
MEM(father) MEM(siz) MEM(root)
For(i,n+1) siz[i]=1,root[i]=1;root[0]=1;
MEM(ch)
}
void maintain(int x)
{
siz[x]=siz[ch[x][0]]+siz[ch[x][1]]+1;
}
void rotate(int x)
{
int y=father[x],kind=ch[y][1]==x;
ch[y][kind]=ch[x][!kind];
if (ch[y][kind]) {
father[ch[y][kind]]=y;
}
father[x]=father[y];
father[y]=x;
ch[x][!kind]=y;
if (root[y])
{
root[x]=1;root[y]=0;
}
else
{
ch[father[x]][ ch[father[x]][1]==y ] = x;
}
maintain(y);maintain(x);
}
void splay(int x)
{
while(!root[x])
{
int y=father[x];
int z=father[y];
if (root[y]) rotate(x);
else if ( (ch[y][1]==x)^(ch[z][1]==y) )
{
rotate(x); rotate(x);
}
else
{
rotate(y); rotate(x);
}
}
}
int access(int x)
{
int y=0;
do
{
splay(x);
if (ch[x][1]) root[ch[x][1]]=1; root[ch[x][1]=y]=0;
maintain(x);
y = x;
x=father [ x];
} while (x) ;
return y;
}
}S;
int main()
{
// freopen("bzoj2002.in","r",stdin);
// freopen(".out","w",stdout);
cin>>n;
S.mem(n);
For(i,n)
{
scanf("%d",&S.father[i]);
S.father[i]+=i;
if (S.father[i]>n) S.father[i]=n+1;
}
cin>>m;
For(i,m)
{
int p;
scanf("%d",&p);
if (p==1)
{
int x;
scanf("%d",&x);x++;
S.access(x);
S.splay(x);
printf("%d\n",S.siz[S.ch[x][0]]);
} else {
int x,k;
scanf("%d%d",&x,&k);x++;
k+=x;
if (k>n) k=n+1;
S.access(x);
S.splay(x);
S.father[S.ch[x][0]]=S.father[x];
S.father[x]=0;
S.root[S.ch[x][0]]=1;
S.ch[x][0]=0;
S.maintain(x);
S.father[x]=k;
}
}
return 0;
}
8.28 删除了没有多余的代码
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (200000+10)
#define MAXM (100000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int n,m;
class Splay
{
public:
int father[MAXN],siz[MAXN];
int ch[MAXN][2];
bool root[MAXN];
void mem(int n)
{
MEM(father) MEM(siz) MEM(root)
For(i,n+1) siz[i]=1,root[i]=1;root[0]=1;
MEM(ch)
}
void maintain(int x)
{
siz[x]=siz[ch[x][0]]+siz[ch[x][1]]+1;
}
void rotate(int x)
{
int y=father[x],kind=ch[y][1]==x;
ch[y][kind]=ch[x][!kind];
if (ch[y][kind]) {
father[ch[y][kind]]=y;
}
father[x]=father[y];
father[y]=x;
ch[x][!kind]=y;
if (root[y])
{
root[x]=1;root[y]=0;
}
else
{
ch[father[x]][ ch[father[x]][1]==y ] = x;
}
maintain(y);maintain(x);
}
void splay(int x)
{
while(!root[x])
{
int y=father[x];
int z=father[y];
if (root[y]) rotate(x);
else if ( (ch[y][1]==x)^(ch[z][1]==y) )
{
rotate(x); rotate(x);
}
else
{
rotate(y); rotate(x);
}
}
}
int access(int x)
{
int y=0;
do
{
splay(x);
if (ch[x][1]) root[ch[x][1]]=1;
ch[x][1]=y;
if (y) root[y]=0;
maintain(x);
y = x;
x=father [x];
} while (x) ;
return y;
}
void cut(int x)
{
access(x);
splay(x);
father[ch[x][0]]=0;
root[ch[x][0]]=1;
ch[x][0]=0;
maintain(x);
}
void join(int x,int w)
{
father[x]=w;
}
}S;
int main()
{
// freopen("bzoj2002.in","r",stdin);
// freopen(".out","w",stdout);
cin>>n;
S.mem(n);
For(i,n)
{
scanf("%d",&S.father[i]);
S.father[i]+=i;
if (S.father[i]>n) S.father[i]=n+1;
}
cin>>m;
For(i,m)
{
int p;
scanf("%d",&p);
if (p==1)
{
int x;
scanf("%d",&x);x++;
S.access(x);
S.splay(x);
printf("%d\n",S.siz[S.ch[x][0]]);
} else {
int x,k;
scanf("%d%d",&x,&k);x++;
k+=x;
if (k>n) k=n+1;
S.cut(x);
S.join(x,k);
}
}
return 0;
}