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HDOJ 5141 LIS again 二分
二分求LIS
对每一个位置为终点的LIS记录开头的最靠右边的值....
LIS again
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 272 Accepted Submission(s): 96
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 ).
S[ i , j ] indicates (
Your program, when given the numeric sequence (
Input
Multi 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
Process to the end of file.
[Technical Specification]
Output
For each case,.output the answer in a single line.
Sample Input
3 1 2 3 2 2 1
Sample Output
1 3
Source
BestCoder Round #21
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long long int LL; const int maxn=100100; int a[maxn],c[maxn],w[maxn],p[maxn],l[maxn]; int len,n; LL ans; void init() { ans=0; len=0; memset(w,-1,sizeof(w)); } int main() { while(scanf("%d",&n)!=EOF) { init(); for(int i=0;i<n;i++) scanf("%d",a+i); len=1; c[0]=a[0]; l[0]=1; p[0]=0; w[0]=0; for(int i=1;i<n;i++) { int j=lower_bound(c,c+len,a[i])-c; c[j]=a[i]; if(j==0) w[j]=i; else w[j]=max(w[j],w[j-1]); p[i]=w[j]; l[i]=j+1; if(j>=len) len++; } bool flag=false; int l1=-1; for(int i=0;i<n;i++) { if(l[i]==len) { flag=true; l1=max(l1,p[i]); } if(flag) { ans+=l1+1; } } cout<<ans<<endl; } return 0; }
HDOJ 5141 LIS again 二分
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