HDU_5141 LIS again[线段树]

传送门：HUD_5141

LIS again

[align=left]Problem Description[/align]
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).
 

[align=left]Input[/align]
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
 

[align=left]Output[/align]
For each case，.output the answer in a single line.
 

[align=left]Sample Input[/align]

3
1 2 3
2
2 1

 

[align=left]Sample Output[/align]

1
3

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|1
using namespace std;
typedef long long LL;
const int N=100010;

int a
,b
;
int dp
;
int sta
,en
;
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;
}