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

2002: [Hnoi2010]Bounce 弹飞绵羊

Time Limit: 10 Sec  Memory Limit: 259 MB
Submit: 5392  Solved: 2846
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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

维护子树的大小，注意下标是从0开始的。

#include <iostream>#include <cstdio>using namespace std;const int N = 2 * 1e5 + 5;#define LS(x) node[(x)].ch[0]#define RS(x) node[(x)].ch[1]int nxt[N];struct LCT {struct Node {int fa, ch[2];bool rev;int size;void init() {fa = 0;rev = ch[0] = ch[1] = 0;size = 1;}} node[N];bool is_root(int x) {return (LS(node[x].fa) != x && RS(node[x].fa) != x);}void down(int x) {if(node[x].rev) {node[LS(x)].rev ^= 1;node[RS(x)].rev ^= 1;swap(LS(x), RS(x));node[x].rev = 0;}}void up(int x) {node[x].size = node[LS(x)].size + node[RS(x)].size + 1;}void rotate(int x, bool kind) {int fx = node[x].fa;int ffx = node[fx].fa;node[fx].ch[!kind] = node[x].ch[kind];node[node[x].ch[kind]].fa = fx;if(!is_root(fx))node[ffx].ch[RS(ffx) == fx] = x;node[x].fa = ffx;node[x].ch[kind] = fx;node[fx].fa = x;up(fx);}int d_st[N], ds_top;void splay(int x) {ds_top = -1;d_st[++ds_top] = x;for(int i = x; !is_root(i); i = node[i].fa)d_st[++ds_top] = node[i].fa;for(int i = ds_top; i >= 0 ; --i) down(d_st[i]);while(!is_root(x)) {int fx = node[x].fa;int ffx = node[fx].fa;bool rot_x = (LS(fx) == x);bool rot_fx = (LS(ffx) == fx);if(is_root(fx)) rotate(x, rot_x);else {if(rot_x == rot_fx) rotate(fx, rot_fx);else rotate(x, rot_x);rotate(x, rot_fx);}}up(x);}void access(int x) {int last = 0;while(x) {splay(x);RS(x) = last;last = x;x = node[x].fa;}}//access(x); splay(x); 后x在根的位置，此时x只有左子树，没有右子树，打翻转标记就是使得左子树变到右子树上。void make_root(int x) {access(x); splay(x);node[x].rev ^= 1;}void cut(int x, int y) {make_root(x);access(y); splay(y);LS(y) = node[x].fa = 0;up(y);}void link(int x, int y) {make_root(x);node[x].fa = y;}void init(int n) {for(int i = 0; i <= n; ++i)node[i].init();node[0].size = 0;}void debug(int n) {for(int i = 1; i <= n; ++i) {cout<<i<<": "<<node[i].size<<' '<<node[i].fa<<' '<<node[i].ch[0]<<' '<<node[i].ch[1]<<endl;}}void op1(int x, int n) {make_root(n + 1);access(x); splay(x);printf("%d\n", node[LS(x)].size);}void op2(int x, int y, int n) {int u = x, v = min(n + 1, x + y), z = nxt[x];cut(u, z);link(u, v);nxt[x] = v;}} lct;int main() {int n;scanf("%d", &n);lct.init(n + 1);for(int i = 1; i <= n; ++i) {int a;scanf("%d", &a);lct.node[i].fa = i + a;if(lct.node[i].fa > n) lct.node[i].fa = n + 1;nxt[i] = lct.node[i].fa;}//lct.debug(n + 1);int m;scanf("%d", &m);for(int i = 0; i < m; ++i) {int op, x, y;scanf("%d", &op);if(op == 1) {scanf("%d", &x);++x;lct.op1(x, n);} else {scanf("%d%d", &x, &y);++x;lct.op2(x, y, n);}//lct.debug(n + 1);}return 0;}