HDOJ question 3564 Another LIS (LIS), hdoj3564

HDOJ question 3564 Another LIS (LIS), hdoj3564
Another LISTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission (s): 1291 Accepted Submission (s): 451


Problem DescriptionThere is a sequence firstly empty. we begin to add number from 1 to N to the sequence, and every time we just add a single number to the sequence at a specific position. now, we want to know length of the LIS (Longest Increasing Subsequence) after every time's add.
InputAn integer T (T <= 10), indicating there are T test cases.
For every test case, an integer N (1 <= N <= 100000) comes first, then there are N numbers, the k-th number Xk means that we add number k at position Xk (0 <= Xk <= k-1 ). see hint for more details.
 
OutputFor the k-th test case, first output "Case # k:" in a separate line, then followed N lines indicating the answer. Output a blank line after every test case.
Sample Input

130 0 2

 
Sample Output

Case #1:112HintIn the sample, we add three numbers to the sequence, and form three sequences.a. 1b. 2 1c. 2 1 3 

 
Authorstandy
Source2010 ACM-ICPC Multi-University Training Contest (13) -- Host by UESTC
Recommendzhouzeyong | We have carefully selected several similar problems for you: 3572 2389 3584 3293 idea: It is to insert a blank space from the left, from 1 ~ N use a line segment tree to record their locations, and then perform LIS on their locations. ac code

#include<stdio.h>#include<string.h>#define max(a,b) (a>b?a:b)int a[100010];int node[100010<<2],d[100010],len,dp[100010];void build(int l,int r,int tr){node[tr]=r-l+1;if(l==r)return;int mid=(l+r)>>1;build(l,mid,tr<<1);build(mid+1,r,tr<<1|1);node[tr]=node[tr<<1]+node[tr<<1|1];}int bin(int x){int l=1,r=len;while(l<=r){int mid=(l+r)>>1;if(x>dp[mid])l=mid+1;elser=mid-1;}return l;}void insert(int pos,int num,int l,int r,int tr){if(l==r){d[num]=l;node[tr]=0;return;}int mid=(l+r)>>1;node[tr]--;if(pos<=node[tr<<1]){insert(pos,num,l,mid,tr<<1);}elseinsert(pos-node[tr<<1],num,mid+1,r,tr<<1|1);}int main(){int t,c=0;scanf("%d",&t);while(t--){int n;scanf("%d",&n);int i;for(i=1;i<=n;i++){scanf("%d",&a[i]);dp[i]=0;}build(1,n,1);for(i=n;i>0;i--){insert(a[i]+1,i,1,n,1);}len=0;/*for(i=1;i<=n;i++){printf("%d\n",d[i]);}*/printf("Case #%d:\n",++c);for(i=1;i<=n;i++){int k=bin(d[i]);len=max(len,k);dp[k]=d[i];printf("%d\n",len);}printf("\n");}}

 

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