HDU_5141 LIS again (lis)

HDU_5141 LIS again (lis)

Portal: HUD_5141

LIS again
Problem DescriptionA numeric sequence of ai is ordered if A1. Let the subsequence of the given numeric sequence ( A1, a2 ,..., AN ) Be any sequence ( Ai1, ai2 ,..., AiK ), Where 1 ≤ i1 . For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, eg. (1, 7), (3, 4, 8) and other others.
S [I, j] indicates ( Ai, ai + 1, ai + 2 ,..., Aj ).
Your program, when given the numeric sequence ( A1, a2 ,..., AN ), Must find the number of pair (I, j) which makes the length of the longest ordered subsequence of S [I, j] equals to the length of the longest ordered subsequence ( A1, a2 ,..., AN ).
InputMulti test cases (about 100), every case occupies two lines, the first line contain n, then second line contain n numbers A1, a2 ,..., AN Separated by exact one space.
Process to the end of file.

[Technical Specification]
1 ≤ n ≤100000
0 ≤ ai ≤ 1000000000
OutputFor each case,. output the answer in a single line.
Sample Input

31 2 322 1

Sample Output

13

Question: Here is a series. The longest ascending sequence is lis, and the number of ranges in which the length of the ascending sequence is equal to lis.

Idea: The line segment tree maintains the longest ascending sequence ending with I, and requires that the start point be right-aligned as much as possible.

Code:

#include
 
  #include
  
   #include
   
    #include#define ls(p) p<<1#define rs(p) p<<1|1using namespace std;typedef long long LL;const int N=100010;int a[N],b[N];int dp[N];int sta[N],en[N];int len,st;struct node{    int l,r;    int len,sta;}tree[N<<2];void pushup(int p){    if(tree[ls(p)].len>tree[rs(p)].len)    {        tree[p].len=tree[ls(p)].len;        tree[p].sta=tree[ls(p)].sta;    }    else if(tree[rs(p)].len>tree[ls(p)].len)    {        tree[p].len=tree[rs(p)].len;        tree[p].sta=tree[rs(p)].sta;    }    else        tree[p].sta=max(tree[ls(p)].sta,tree[rs(p)].sta);}void build(int p,int l,int r){    tree[p].l=l;tree[p].r=r;    tree[p].len=-1;    tree[p].sta=-1;    if(l==r)        return;    int m=(l+r)>>1;    build(ls(p),l,m);    build(rs(p),m+1,r);}void update(int p,int pos,int len,int st){    if(tree[p].l==tree[p].r)    {        if(tree[p].len==len&&tree[p].sta
    
     >1;    if(pos<=m) update(ls(p),pos,len,st);    else update(rs(p),pos,len,st);    pushup(p);}void query(int p,int l,int r){    if(l<=tree[p].l&&tree[p].r<=r)    {        if(tree[p].len>len)        {            len=tree[p].len;            st=tree[p].sta;        }        else if(tree[p].len==len&&st
     
      >1;    if(l<=m) query(ls(p),l,r);    if(r>m) query(rs(p),l,r);}int main(){    int n;    while(scanf("%d",&n)==1)    {        LL ret=0;        int cnt;        for(int i=1;i<=n;i++)        {            scanf("%d",a+i);            b[i]=a[i];        }        if(n==1)        {            printf("1\n");            continue;        }        sort(b+1,b+1+n);        cnt=unique(b+1,b+1+n)-(b+1);        dp[1]=sta[1]=1;        build(1,1,cnt);        int mpp;        int lest=-1;        for(int i=1;i<=n;i++)        {            len=st=-1;            mpp=lower_bound(b+1,b+1+cnt,a[i])-b;            if(mpp==1)            {                dp[i]=1;                sta[i]=i;            }            else                query(1,1,mpp-1);            if(st==-1)            {                dp[i]=1;                sta[i]=i;            }            else            {                dp[i]=len+1;                sta[i]=st;            }            //cout<
      
       =1;i--) { if(dp[i]==lest) { en[i]=last-1; last=i; } } for (int i=1;i<=n;++i) { if (dp[i]==lest) ret+=(LL)sta[i]*(LL)(en[i]-i+1); } printf("%I64d\n", ret); } return 0;}