hdu-6406-dp+ST表
Taotao Picks Apples
Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1446 Accepted Submission(s): 449
When Taotao picks apples, Taotao scans these apples from the first one to the last one. If the current apple is the first apple, or it is strictly higher than the previously picked one, then Taotao will pick this apple; otherwise, he will not pick.
Given the heights of these apples h1,h2,⋯,hn, you are required to answer some independent queries. Each query is two integers p,q, which asks the number of apples Taotao would pick, if the height of the p-th apple were q (instead of hp). Can you answer all these queries?
Each test case begins with a line of two integers n,m (1≤n,m≤105), denoting the number of apples and the number of queries. It is then followed by a single line of n integers h1,h2,⋯,hn (1≤hi≤109), denoting the heights of the apples. The next m lines give the queries. Each of these m lines contains two integers p (1≤p≤n) and q (1≤q≤109), as described in the problem statement.
5 3
1 2 3 4 4
1 5
5 5
2 3
#include <iostream>
#include<cmath>
using namespace std;
int N,a[];
int f[][],dp[][];
void init(){
for(int i=;i<=N;++i)f[i][]=i;
for(int k=;(<<k)<=N;++k){
for(int i=;i+(<<k)-<=N;++i){
if(a[f[i][k-]]>=a[f[i+(<<(k-))][k-]]) f[i][k]=f[i][k-];
else f[i][k]=f[i+(<<(k-))][k-];
}
}
}
int query(int L,int R){
if(R<L)return ;
int k=;
while((<<(k+))-<=R-L) k++;
if(a[f[L][k]]>=a[f[R-(<<k)+][k]]) return f[L][k];
else return f[R-(<<k)+][k];
}
int main() {
int t,n,m,i,p,q;
cin>>t;
while(t--){
scanf("%d%d",&n,&m);
for(i=;i<=n;++i)scanf("%d",a+i);
N=n,init();
dp[][]=;
int maxn=;
for(i=;i<=n;++i){
if(a[i]>a[maxn]){
dp[][i]=dp[][maxn]+;
maxn=i;
}
else dp[][i]=-;
} dp[][n]=;
for(i=n-;i>=;--i){
int l=i+,r=n;
while(l<r){
int mid=l+(r-l)/;
if(a[query(l,mid)]>a[i]) r=mid;
else l=mid+;
}
dp[][i]=a[r]>a[i]?dp[][l]+:;
}
while(m--){
scanf("%d%d",&p,&q);
int x=query(,p-),xx=a[x],ans=dp[][x];
if(q>a[x]) xx=q,ans++;
int l=p+,r=n;
while(l<r){
int mid=l+(r-l)/;
if(a[query(p+,mid)]>xx)r=mid;
else l=mid+;
}
if(a[l]>xx) ans+=dp[][l];
cout<<ans<<endl;
}
}
return ;
}
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