HDU_5141 LIS again[线段树]

传送门：HUD_5141

LIS again

Problem Description

A numeric sequence of ai is ordered if a1<a2<…<aN. Let the subsequence of the given numeric sequence (a1,a2,…,aN) be any sequence (ai1,ai2,…,aiK), where 1≤i1<i2<…<iK≤N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, eg. (1, 7), (3, 4, 8) and many 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 of (a1,a2,…,aN).

 

Input

Multi test cases (about 100), every case occupies two lines, the first line contain n, then second line contain n numbersa1,a2,…,aN separated by exact one space.
Process to the end of file.

[Technical Specification]
1≤n≤100000
0≤ai≤1000000000

 

Output

For each case，.output the answer in a single line.

 

Sample Input

31 2 322 1

 

Sample Output

13

题意：给你一串数列，其最长递增序列为lis，统计递增序列长度等于lis的区间的数目。

思路：线段树维护以i结尾的最长递增序列，且要求起点尽量靠右。

代码：

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>#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<st)            tree[p].sta=st;        else if(tree[p].len<len)        {            tree[p].len=len;            tree[p].sta=st;        }        return;    }    int m=(tree[p].l+tree[p].r)>>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<tree[p].sta)            st=tree[p].sta;        return;    }    int m=(tree[p].l+tree[p].r)>>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<<mpp<<dp[i]<<sta[i]<<"--\n";            update(1,mpp,dp[i],sta[i]);            lest=max(lest,dp[i]);        }        int last=n+1;        for(int i=n;i>=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;}