[BZOJ2002] Bounce 弹飞绵羊(分块)

题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=2002

分块，每块中的每一个点除了自己能跳跃的距离外，还要额外维护两个值：to(i)：i跳出本块到其他块的id，st(i)：i跳出本块所需要的次数。

每次暴力更新本块内的所有值，查询的时候一直走就好。

#include <bits/stdc++.h>
using namespace std;
#define fr first
#define sc second
#define cl clear
#define BUG puts("here!!!")
#define W(a) while(a--)
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rll(a) scanf("%lld", &a)
#define Rs(a) scanf("%s", a)
#define Cin(a) cin >> a
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Clr(a, in) memset((a), (in), sizeof(a))
#define Full(a) memset((a), 0x7f7f7f, sizeof(a))
#define lrt rt << 1
#define rrt rt << 1 | 1
#define pi 3.14159265359
#define RT return
#define lowbit(in) in & (-in)
#define onenum(in) __builtin_popcount(in)
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<int, int> pii;
typedef pair<string, int> psi;
typedef pair<LL, LL> pll;
typedef map<string, int> msi;
typedef vector<int> vi;
typedef vector<LL> vl;
typedef vector<vl> vvl;
typedef vector<bool> vb;

const int maxn = 200200;
int n, q, sz, cmd;
int st[maxn], to[maxn], k[maxn], be[maxn];

void update(int pos, int val) {
    k[pos] = val;
    for(int i = pos; i >= be[pos] * sz; i--) {
        if(i + k[i] > n - 1) {
            st[i] = 1;
            to[i] = -1;
        }
        else {
            if(be[i] == be[i+k[i]]) {
                st[i] = st[i+k[i]] + 1;
                to[i] = to[i+k[i]];
            }
            else {
                st[i] = 1;
                to[i] = i + k[i];
            }
        }
    }
}

int query(int pos) {
    int ret = 0;
    while(pos != -1) {
        ret += st[pos];
        pos = to[pos];
    }
    return ret;
}

signed main() {
    // FRead();
    int x, y;
    while(~Rint(n)) {
        sz = sqrt(n);
        Cls(st); Cls(to);
        Rep(i, n) {
            Rint(k[i]);
            be[i] = i / sz;
        }
        for(int i = n - 1; i >= 0; i--) update(i, k[i]);
        Rint(q);
        W(q) {
            Rint(cmd);
            if(cmd == 1) {
                Rint(x);
                printf("%d
", query(x));
            }
            else {
                Rint(x); Rint(y);
                update(x, y);
            }
        }
    }
    RT 0;
}