HDU 4718 The LCIS on the Tree

For a sequence S 1, S 2, ... , S N, and a pair of integers (i, j), if 1 <= i <= j <= N and S i < S i+1 < S i+2 < ... < S j-1 < S j , then the
sequence S i, S i+1, ... , S jis a CIS (Continuous Increasing Subsequence). The longest CIS of a sequence is called the LCIS (Longest Continuous Increasing Subsequence). 

Now we consider a tree rooted at node 1. Nodes have values. We have Q queries, each with two nodes u and v. You have to find the shortest path from u to v. And write down each nodes' value on the path, from u to v, inclusive. Then you will get a sequence, and
please show us the length of its LCIS. 

InputThe first line has a number T (T <= 10) , indicating the number of test cases. 

For each test case, the first line is a number N (N <= 10 5), the number of nodes in the tree. 

The second line comes with N numbers v1, v2, v3 ... , v N, describing the value of node 1 to node N. (1 <= v i <= 10 9) 

The third line comes with N - 1 numbers p 2, p 3, p 4 ... , p N, describing the father nodes of node 2 to node N. Node 1 is the root and will have no father. 

Then comes a number Q, it is the number of queries. (Q <= 10 5) 

For next Q lines, each with two numbers u and v. As described above.
OutputFor test case X, output "Case #X:" at the first line. 

Then output Q lines, each with an answer to the query. 

There should be a blank line *BETWEEN* each test case.
Sample Input

1
5
1 2 3 4 5
1 1 3 3
3
1 5
4 5
2 5

Sample Output

Case #1:
3
2

3
最长连续上升子串的树上版本，考虑到线性的可以使用线段树，树上的话，使用树链剖分拆成线性的即可。这题的数据貌似比较水，看到暴力的程序竟然能跑过去。
#include<map>
#include<set>
#include<ctime>
#include<cmath>
#include<stack>
#include<queue>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define ms(x,y) memset(x,y,sizeof(x))
#define rep(i,j,k) for(int i=j;i<=k;i++)
#define per(i,j,k) for(int i=j;i>=k;i--)
#define loop(i,j,k) for (int i=j;i!=-1;i=k[i])
#define inone(x) scanf("%d",&x)
#define intwo(x,y) scanf("%d%d",&x,&y)
#define inthr(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define infou(x,y,z,p) scanf("%d%d%d%d",&x,&y,&z,&p)
#define lson x<<1,l,mid
#define rson x<<1|1,mid+1,r
#define mp(i,j) make_pair(i,j)
#define ff first
#define ss second
typedef long long LL;
typedef pair<int, int> pii;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int N = 1e5 + 10;
const double eps = 1e-8;
int T, cas = 1, n, m, x, y;
int v
, fa
, ft
, nt
, u
, sz;
int mx
, ct
, top
, dep
;
int g
, f
, tot;

struct point
{
int l, r, ll, rr, mx, len;
point(int l = 0, int ll = 0, int r = 0, int rr = 0, int mx = 0, int len = 0)
{
this->l = l; this->ll = ll;
this->r = r; this->rr = rr;
this->mx = mx; this->len = len;
}
}p[N << 2];

point merge(point a, point b)
{
point c;
c.l = a.l; c.r = b.r; c.len = a.len + b.len;
c.ll = a.ll == a.len && a.r < b.l ? a.len + b.ll : a.ll;
c.rr = b.rr == b.len && b.l > a.r ? b.len + a.rr : b.rr;
c.mx = max(max(a.mx, b.mx), a.r < b.l ? a.rr + b.ll : 0);
return c;
}

void dfs(int x, int d)
{
mx[x] = 0; dep[x] = d;
loop(i, ft[x], nt)
{
dfs(u[i], d + 1);
mx[x] = ct[u[i]] > ct[mx[x]] ? u[i] : mx[x];
}
}

void Dfs(int x, int tp)
{
g[++tot] = x; f[x] = tot;
top[x] = tp ? top[fa[x]] : x;
if (mx[x]) Dfs(x, 1);
loop(i, ft[x], nt)
{
if (u[i] == mx[x]) continue;
Dfs(u[i], 0);
}
}

void build(int x, int l, int r)
{
if (l == r) { p[x] = point(v[g[r]], 1, v[g[r]], 1, 1, 1); return; }
int mid = l + r >> 1;
build(lson); build(rson);
p[x] = merge(p[x << 1], p[x << 1 | 1]);
}

point get(int x, int l, int r, int ll, int rr)
{
if (ll <= l&&r <= rr) return p[x];
int mid = l + r >> 1;
if (rr <= mid) return get(lson, ll, rr);
if (ll > mid) return get(rson, ll, rr);
return merge(get(lson, ll, rr), get(rson, ll, rr));
}

int main()
{
for (inone(T); T--; cas++)
{
inone(n); ct[0] = sz = 0;
rep(i, 1, n) inone(v[i]);
rep(i, 1, n) ft[i] = -1;
rep(i, 2, n)
{
inone(fa[i]);
u[sz] = i; nt[sz] = ft[fa[i]]; ft[fa[i]] = sz++;
}
dfs(1, 1); Dfs(1, tot = 0);
printf("Case #%d:\n", cas);
build(1, 1, n);
for (inone(m); m--;)
{
intwo(x, y);
point l = point(0, 0, 0, 0, 0, 0), r = point(INF, 0, INF, 0, 0, 0);
while (top[x] != top[y])
{
if (dep[x] > dep[y])
{
l = merge(l, get(1, 1, n, f[top[x]], f[x]));
x = fa[top[x]];
}
else
{
r = merge(get(1, 1, n, f[top[y]], f[y]), r);
y = fa[top[y]];
}
}
if (dep[x] > dep[y]) swap(x, y);
l = merge(l, merge(get(1, 1, n, f[x], f[y]), r));
printf("%d\n", l.mx);
}
if (T) putchar(10);
}
return 0;
}