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. 

Input

The 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.

Output

For 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

151 2 3 4 51 1 3 331 54 52 5

Sample Output

Case #1:32
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[N], fa[N], ft[N], nt[N], u[N], sz;int mx[N], ct[N], top[N], dep[N];int g[N], f[N], 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;}