The LCIS on the Tree
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1615 Accepted Submission(s): 457
题目链接
https://vjudge.net/problem/UVA-12655
Problem Description
For a sequence S1, S2, ... , SN, and a pair of integers (i, j), if 1 <= i <= j <= N and Si < Si+1 < Si+2 < ... < Sj-1 < Sj , then the sequence Si, Si+1, ... , Sj is 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 <= 105), the number of nodes in the tree.
The second line comes with N numbers v1, v2, v3 ... , vN, describing the value of node 1 to node N. (1 <= vi <= 109)
The third line comes with N - 1 numbers p2, p3, p4 ... , pN, 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 <= 105)
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
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
题意
在一棵树上，每次求一条路径最长连续上升子序列。
题解
一开始看错题，以为求最长上升子序列，然后想了好久毫无思路，一看题解发现原来我看错题了。
连续的话就不难了。
树链剖分，然后线段树记录一下该区间最长上升子序列，左端点开始的最长上升子序列长度，右端点。。。
反正就是考虑如果知道左子树和右子树的信息，如何合并。显然合并就是中间可能延长，其他的都一样。
但是在树上，x-->lca(x,y) 和 lca(x,y)-->是不一样的，x-->lca(x,y)需要求最长下降子序列，然后就是一对细节需要处理，具体看代码吧。
代码
1 #include<bits/stdc++.h>
2 using namespace std;
3 #define ll long long
4 #define N 500050
5 struct Tree{int l,r,lb,rb,lI,rI,lD,rD,Imx,Dmx;}tr[N<<2];
6 struct Edge{int from,to,s;}edges[N<<1];
7 int n,m,cas,w[N];
8 int tot,last[N];
9 int cnt,fa[N],dp[N],size[N],son[N],rk[N],kth[N],top[N];
10 template<typename T>void read(T&x)
11 {
12 ll k=0; char c=getchar();
13 x=0;
14 while(!isdigit(c)&&c!=EOF)k^=c==‘-‘,c=getchar();
15 if (c==EOF)exit(0);
16 while(isdigit(c))x=x*10+c-‘0‘,c=getchar();
17 x=k?-x:x;
18 }
19 void read_char(char &c)
20 {while(!isalpha(c=getchar())&&c!=EOF);}
21 void AddEdge(int x,int y)
22 {
23 edges[++tot]=Edge{x,y,last[x]};
24 last[x]=tot;
25 }
26 void dfs1(int x,int pre)
27 {
28 fa[x]=pre;
29 dp[x]=dp[pre]+1;
30 size[x]=1;
31 son[x]=0;
32 for(int i=last[x];i;i=edges[i].s)
33 {
34 Edge &e=edges[i];
35 if (e.to==pre)continue;
36 dfs1(e.to,x);
37 size[x]+=size[e.to];
38 if(size[e.to]>size[son[x]])son[x]=e.to;
39 }
40 }
41 void dfs2(int x,int y)
42 {
43 rk[x]=++cnt;
44 kth[cnt]=x;
45 top[x]=y;
46 if (son[x]==0)return;
47 dfs2(son[x],y);
48 for(int i=last[x];i;i=edges[i].s)
49 {
50 Edge &e=edges[i];
51 if (e.to==fa[x]||e.to==son[x])continue;
52 dfs2(e.to,e.to);
53 }
54 }
55 void init(Tree &a){a=Tree{0,0,0,0,0,0,0,0,0,0};}
56 template<typename T>void he(T &c,T a, T b)
57 {
58 if (a.Imx==0){c=b;return;}
59 if (b.Imx==0){c=a;return;}
60 c.l=a.l; c.r=b.r;
61 int lena=a.r-a.l+1,lenb=b.r-b.l+1;
62 c.lb=a.lb; c.rb=b.rb;
63 c.lI=lena==a.lI&&a.rb<b.lb?lena+b.lI:a.lI;
64 c.lD=lena==a.lD&&a.rb>b.lb?lena+b.lD:a.lD;
65 c.rI=lenb==b.rI&&a.rb<b.lb?lenb+a.rI:b.rI;
66 c.rD=lenb==b.rD&&a.rb>b.lb?lenb+a.rD:b.rD;
67 c.Imx=max(a.Imx,b.Imx);
68 c.Imx=max(c.Imx,a.rb<b.lb?a.rI+b.lI:0);
69 c.Dmx=max(a.Dmx,b.Dmx);
70 c.Dmx=max(c.Dmx,a.rb>b.lb?a.rD+b.lD:0);
71 }
72 void bt(int x,int l,int r)
73 {
74 tr[x].l=l; tr[x].r=r;
75 if (l==r)
76 {
77 tr[x]={l,r,w[kth[l]],w[kth[l]],1,1,1,1,1,1};
78 return;
79 }
80 int mid=(l+r)>>1;
81 bt(x<<1,l,mid);
82 bt(x<<1|1,mid+1,r);
83 he(tr[x],tr[x<<1],tr[x<<1|1]);
84 }
85 Tree query(int x,int l,int r)
86 {
87 if (l<=tr[x].l&&tr[x].r<=r)
88 return tr[x];
89 int mid=(tr[x].l+tr[x].r)>>1;
90 Tree tp1,tp2,tp; init(tp1); init(tp2);
91 if (l<=mid)tp1=query(x<<1,l,r);
92 if (mid<r)tp2=query(x<<1|1,l,r);
93 he(tp,tp1,tp2);
94 return tp;
95 }
96 int get_max(int x,int y)
97 {
98 int fx=top[x],fy=top[y],ans=0;
99 Tree tpx,tpy,tp;
100 init(tpx); init(tpy);
101 while(fx!=fy)
102 {
103 if (dp[fx]>dp[fy])
104 {
105 tp=query(1,rk[fx],rk[x]);
106 he(tpx,tp,tpx);
107 x=fa[fx]; fx=top[x];
108 }
109 else
110 {
111 tp=query(1,rk[fy],rk[y]);
112 he(tpy,tp,tpy);
113 y=fa[fy]; fy=top[y];
114 }
115 }
116 if (dp[x]>dp[y])
117 {
118 tp=query(1,rk[y],rk[x]);
119 he(tpx,tp,tpx);
120 }
121 else
122 {
123 tp=query(1,rk[x],rk[y]);
124 he(tpy,tp,tpy);
125 }
126 ans=max(tpx.Dmx,tpy.Imx);
127 ans=max(ans,tpx.lb<tpy.lb?tpx.lD+tpy.lI:0);
128 return ans;
129 }
130 void work()
131 {
132 if (cas)printf("\n");
133 printf("Case #%d:\n",++cas);
134 read(n);
135 for(int i=1;i<=n;i++)read(w[i]);
136 for(int i=2;i<=n;i++)
137 {
138 int x;
139 read(x);
140 AddEdge(x,i);
141 }
142 dfs1(1,0);
143 dfs2(1,1);
144 bt(1,1,n);
145 read(m);
146 for(int i=1;i<=m;i++)
147 {
148 int x,y;
149 read(x); read(y);
150 printf("%d\n",get_max(x,y));
151 }
152 }
153 void clear()
154 {
155 cnt=0; tot=0;
156 memset(last,0,sizeof(last));
157 }
158 int main()
159 {
160 #ifndef ONLINE_JUDGE
161 freopen("aa.in","r",stdin);
162 #endif
163 int q;
164 read(q);
165 while(q--)
166 {
167 clear();
168 work();
169 }
170 }
原文地址：https://www.cnblogs.com/mmmqqdd/p/10793607.html