bzoj_p2002 [Hnoi2010]bounce Fly Sheep (LCT)

Bzoj Portal
Time Limit:10 Sec Memory limit:259 MB
submit:6636 solved:3478
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Description
One day, Lostmonkey invented a super-elastic device, in order to show off in front of his sheep friends, he invited the little sheep to play a game. At the beginning of the game, Lostmonkey on the ground along a straight line to put n devices, each device set the initial elastic coefficient ki, when the sheep reached the first device, it will bounce Ki step back to the first I+ki device, if there is no i+ki device, then the sheep were bounced. The sheep wanted to know that when it started from the device I, it was bounced a few times. In order to make the game more interesting, Lostmonkey can modify the elastic coefficient of an elastic device, and the elastic coefficient is a positive integer at any time.

Input
The first line contains an integer n, which indicates that there are n devices on the ground, the device is numbered from 0 to N-1, and the next line has n positive integers, followed by the initial elastic coefficients of the n units. The third line has a positive integer m, the next m row each line has at least two number I, J, if I=1, you want to output from J is bounced several times after being bounced, if i=2 will also enter a positive integer k, indicating that the coefficient of the J-Elastic device is modified to K. For data n,m<=10000 of 20%, for 100% of data n<=200000,m<=100000

Output
For each i=1, you will output a required number of steps, one row.

Sample Input
4
1 2 1 1
3
1 1
2 1 1
1 1

Sample Output
2
3

HINT

Source

The first LCT, because the LCT play is not complete, so there is no first hair template.
The idea is very simple, set up a virtual node, all the last pop-up nodes are connected, each time you and the virtual node is located in the root of the splay to u, output u number of child nodes;
for modification is to cut the edge of U to F[u] and then connect u to the new parent node

#include <cstdio> #include <iostream> using namespace std;
    #define N 200005 struct linkcuttree{int n,m;int f[n],ch[n][2],s[n],a[n]; inline bool Is_root (int o) {return (!f[o]) | |
    (O!=ch[f[o]][0]&&o!=ch[f[o]][1]);}
    inline void push (int o) {s[o]=s[ch[o][0]]+s[ch[o][1]]+1;}
        inline void rot (int o) {int p=f[o],k=f[p],d=ch[p][0]==o;
        if (!is_root (p)) ch[k][ch[k][1]==p]=o;
        F[o]=k,f[ch[o][d]]=p;
        Ch[p][d^1]=ch[o][d],ch[o][d]=p;
    F[p]=o;push (p), push (O);
            } inline void splay (int o) {while (!is_root (o)) {int p=f[o],k=f[p];
            if (Is_root (f[o))) rot (o);
            else if ((ch[p][1]==o) = = (ch[k][1]==p)) Rot (p), rot (o);
        else Rot (o), rot (o);
    } push (O);
    } inline void Access (int o,int son=0) {for (; O;o=f[son=o]) splay (o), Ch[o][1]=son,push (o); } inline int in (int x=0,char Ch=getchar ()) {while (ch> ' 9 ' | |
        ch< ' 0 ') Ch=getchar (); while (ch>=' 0 ' &&ch<= ' 9 ') x=x*10+ch-' 0 ', Ch=getchar ();
    return x;
        } void Init () {n=in (); int opt,u,v;
        for (int i=1;i<=n;i++) a[i]=in (), s[i]=1;s[n+1]=1;
        for (int i=1;i<=n;i++) f[i]= (I+a[i]>n+1?n+1:i+a[i]);
        M=in ();
            while (m--) {opt=in (), U=in (); u++;
            if (opt==1) {Access (U), splay (U), printf ("%d\n", s[u]-1);
                } else{v=in ();
                Access (U), splay (U), f[ch[u][0]]=f[u];
                Ch[u][0]=0,f[u]=0;push (U);
            F[u]= (V+U&GT;N+1?N+1:U+V);
}}}}LCT; int main () {lct.init (); return 0;}