【Vijos】lxhgww的奇思妙想

题面

题解

求$k$级祖先孙子

为什么要用长链剖分啊???

倍增并没有慢多少。。。

其实是我不会

长链剖分做这道题还是看这位巨佬的吧。

代码

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#include<bits/stdc++.h>
#define RG register
#define file(x) freopen(#x".in", "r", stdin);freopen(#x".out", "w", stdout);
#define clear(x, y) memset(x, y, sizeof(x));
using namespace std;

inline int read()
{
int data = 0, w = 1;
char ch = getchar();
while(ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if(ch == '-') w = -1, ch = getchar();
while(ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = getchar();
return data*w;
}

const int maxn(300010), Log(21);
struct edge { int next, to; } e[maxn << 1];
int head[maxn], e_num, n, m, f[maxn][Log];
inline void add_edge(int from, int to)
{
e[++e_num] = (edge) {head[from], to}; head[from] = e_num;
e[++e_num] = (edge) {head[to], from}; head[to] = e_num;
}

void dfs(int x)
{
for(RG int i = 1; i <= 20; i++) f[x][i] = f[f[x][i - 1]][i - 1];
for(RG int i = head[x]; i; i = e[i].next)
{
int to = e[i].to;
if(to == f[x][0]) continue;
f[to][0] = x; dfs(to);
}
}

inline int Query(int x, int k)
{
for(RG int i = 20; ~i; i--) if((k >> i) & 1) x = f[x][i];
return x;
}

int main()
{
n = read();
for(RG int i = 1; i < n; i++) add_edge(read(), read());
m = read(); int lastans = 0; dfs(1);
while(m--)
{
int a = read() ^ lastans, b = read() ^ lastans;
printf("%d\n", lastans = Query(a, b));
}
return 0;
}
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