预处理第j个点跳2^i次方所能到达某个点时,
预处理第j个点跳2^(i-1) 与 第j个点跳2^(i-1)次方时所在的点 跳2^(i-1) 的异或(你还是看下面的代码吧)
```cpp
for(int i=1; i<=depth; i++)
for(int j=1; j<=n; j++)
{
f[j][i]=f[f[j][i-1]][i-1];
val[j][i]=val[j][i-1]^val[f[j][i-1]][i-1];
}
```
by OIer991215 @ 2017-06-10 00:12:06
@ chauchat
by OIer991215 @ 2017-06-10 00:12:37
@chauchat
by OIer991215 @ 2017-06-10 00:12:55
不用倍增啊,一遍dfs之后 O(1) 就能回答每组询问
by 金爷爷哈哈 @ 2017-10-03 10:57:28