题意简述
有n个点,第i个点有一个ki,表示到达i这个点后可以到i + ki这个点
支持修改ki和询问一点走几次能走出所有点两个操作
题解思路
分块,
对于每个点,维护它走到下一块所经过的点数,它走到下一块到的店的编号
每次修改只会对这块左端点到这个点产生影响
代码
#include <cmath>
#include <cstdio>
using namespace std;
struct Point{
int k, s, x, num;
}p[200010];
int n, m, len, ii, ans;
void update(int x)
{
int xx = x + p[x].k;
if (xx >= n) p[x].s = 1, p[x].x = -1;
else if (p[x].num != p[xx].num) p[x].s = 1, p[x].x = xx;
else p[x].s = p[xx].s + 1, p[x].x = p[xx].x;
}
int main()
{
scanf("%d", &n);
len = sqrt(n);
for (register int i = 0; i < n; ++i)
{
scanf("%d", &p[i].k);
p[i].num = i / len;
}
for (register int i = n - 1; i + 1; --i)
update(i);
scanf("%d", &m);
for (register int i = 1; i <= m; ++i)
{
int opt;
scanf("%d", &opt);
if (opt == 1)
{
int cnt = 0;
scanf("%d", &ii);
for (ans = 0; ii != -1; ans += p[ii].s, ii = p[ii].x)
++cnt;
printf("%d
", ans);
}
else
{
scanf("%d", &ii);
scanf("%d", &p[ii].k);
for (register int j = ii; j >= 0 && p[j].num == p[ii].num; --j)
update(j);
}
}
}