Problem Description
During his six grade summer vacation, xiaoxin got lots of watermelon candies from his leader when he did his internship at Tencent. Each watermelon candy has it‘s sweetness which denoted by an integer number.
xiaoxin is very smart since he was a child. He arrange these candies in a line and at each time before eating candies, he selects three continuous watermelon candies from a specific range [L, R] to eat and the chosen triplet must satisfies:
if he chooses a triplet (ai,aj,ak) then:
1. j=i+1,k=j+1
2. ai≤aj≤ak
Your task is to calculate how many different ways xiaoxin can choose a triplet in range [L, R]?
two triplets (a0,a1,a2) and (b0,b1,b2) are thought as different if and only if:
a0≠b0 or a1≠b1 or a2≠b2
Input
This problem has multi test cases. First line contains a single integer T(T≤10) which represents the number of test cases.
For each test case, the first line contains a single integer n(1≤n≤200,000)which represents number of watermelon candies and the following line contains n integer numbers which are given in the order same with xiaoxin arranged them from left to right.
The third line is an integer Q(1≤200,000) which is the number of queries. In the following Q lines, each line contains two space seperated integers l,r(1≤l≤r≤n) which represents the range [l, r].
Output
For each query, print an integer which represents the number of ways xiaoxin can choose a triplet.
Sample Input
1 5 1 2 3 4 5 3 1 3 1 4 1 5
Sample Output
1 2 3
这道题额，对于每个点建一棵线段树，表示以它为左端点的所有区间……口胡不清。
1 #include <iostream>
2 #include <cstring>
3 #include <cstdio>
4 #include <algorithm>
5 using namespace std;
6 const int maxn=200010;
7 int a[maxn],Hash[maxn],ok[maxn];
8 int pre[maxn],nxt[maxn],head[maxn],fa[maxn],son[maxn];
9 int sum[maxn*30],ch[maxn*30][2],rt[maxn],cnt;
10 void Insert(int pre,int &rt,int l,int r,int g,int d){
11 rt=++cnt;
12 ch[rt][0]=ch[pre][0];
13 ch[rt][1]=ch[pre][1];
14 sum[rt]=sum[pre]+d;
15 if(l==r)return;
16 int mid=(l+r)>>1;
17 if(mid>=g)Insert(ch[pre][0],ch[rt][0],l,mid,g,d);
18 else Insert(ch[pre][1],ch[rt][1],mid+1,r,g,d);
19 }
20
21 int Query(int rt,int l,int r,int a,int b){
22 if(l>=a&&r<=b)return sum[rt];
23 int mid=(l+r)>>1,ret=0;
24 if(mid>=a)ret=Query(ch[rt][0],l,mid,a,b);
25 if(mid<b)ret+=Query(ch[rt][1],mid+1,r,a,b);
26 return ret;
27 }
28 void Init(){
29 memset(rt,0,sizeof(rt));
30 memset(head,0,sizeof(head));
31 memset(pre,0,sizeof(pre));
32 memset(son,0,sizeof(son));
33 memset(nxt,0,sizeof(nxt));
34 memset(fa,0,sizeof(fa));
35 memset(sum,0,sizeof(sum));
36 memset(ok,0,sizeof(ok));cnt=0;
37 }
38 int main(){
39 int T,Q,n,l,r;a[0]=-1;
40 scanf("%d",&T);
41 while(T--){
42 Init();
43 scanf("%d",&n);
44 for(int i=1;i<=n;i++)
45 scanf("%d",&a[i]);
46 for(int i=1;i<=n;i++)
47 Hash[i]=a[i];
48 sort(Hash+1,Hash+n+1);
49 for(int i=1;i<=n;i++)
50 a[i]=lower_bound(Hash+1,Hash+n+1,a[i])-Hash;
51 for(int i=2;i<n;i++)
52 if(a[i-1]<=a[i]&&a[i]<=a[i+1])
53 ok[i]=1;
54 for(int i=1;i<=n;i++){
55 pre[i]=head[a[i]];
56 nxt[pre[i]]=i;
57 head[a[i]]=i;
58 }
59 for(int i=2;i<n;i++)
60 if(ok[i]){
61 int j=pre[i];
62 while(j&&(a[j-1]!=a[i-1]||a[j+1]!=a[i+1])){j=pre[j];}
63 fa[i]=j;son[j]=i;
64 }
65 for(int i=2;i<n;i++){
66 if(!fa[i]&&ok[i])
67 Insert(rt[1],rt[1],1,n,i,1);
68 }
69
70 for(int i=2;i<n;i++){
71 if(!ok[i]){
72 rt[i]=rt[i-1];
73 continue;
74 }
75 Insert(rt[i-1],rt[i],1,n,i,-1);
76 if(son[i])Insert(rt[i],rt[i],1,n,son[i],1);
77 }
78 scanf("%d",&Q);
79 while(Q--){
80 scanf("%d%d",&l,&r);
81 printf("%d\n",r-1>=l?Query(rt[l],1,n,1,r-1):0);
82 }
83 }
84 return 0;
85 }