hdu5748 Bellovin(LIS lower_bound的使用)
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Bellovin
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1103 Accepted Submission(s): 498
Problem Description
Peter has a sequence a1,a2,...,an and he define a function on the sequence -- F(a1,a2,...,an)=(f1,f2,...,fn) , where fi is the length of the longest increasing subsequence ending with ai .
Peter would like to find another sequenceb1,b2,...,bn in such a manner that F(a1,a2,...,an) equals to F(b1,b2,...,bn) . Among all the possible sequences consisting of only positive integers, Peter wants the lexicographically smallest one.
The sequencea1,a2,...,an is lexicographically smaller than sequence b1,b2,...,bn , if there is such number i from 1 to n , that ak=bk for 1≤k<i and ai<bi .
Peter would like to find another sequence
The sequence
Input
There are multiple test cases. The first line of input contains an integerT , indicating the number of test cases. For each test case:
The first contains an integern (1≤n≤100000) -- the length of the sequence. The second line contains n integers a1,a2,...,an (1≤ai≤109) .
The first contains an integer
Output
For each test case, output n integers b1,b2,...,bn (1≤bi≤109) denoting the lexicographically smallest sequence.
Sample Input
311055 4 3 2 131 3 5
Sample Output
11 1 1 1 11 2 3
题意:求每位的最长上升子序列是多长
用两个辅助数组来进行实现,初始时b[N]中都为很大的数,每次判断第i个数往序列中第一个>=b[i]的地方插入b[i]
#include<cstdio>#include<cstring>#include<algorithm>#define N 100010#define INF 0x3f3f3f3fusing namespace std;int a[N];//保存序列 int b[N];//保存最长子序列 int dp[N];//第i为的最长子序列是多长 int main(){int t;scanf("%d",&t);while(t--){int n;scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d",&a[i]);dp[i]=0; //初始化 b[i]=INF;}for(int i=1;i<=n;i++){int k=lower_bound(b+1,b+n+1,a[i])-b; dp[i]=k;b[k]=a[i];}for(int i=1;i<n;i++) printf("%d ",dp[i]);printf("%d\n",dp[n]);}return 0; }
#include<cstdio> #include<cstring> #include<algorithm> using namespace std;int a[100010];int dp[100010];int c[100010];int main(){int t;scanf("%d",&t);while(t--){int n;scanf("%d",&n);for(int i=0;i<n;i++){scanf("%d",&a[i]);dp[i]=10000000010;} //fill(dp,dp+n,1000000100);for(int i=0;i<n;i++){*lower_bound(dp,dp+n,a[i])=a[i];c[i]=upper_bound(dp,dp+n,a[i])-dp;}for(int i=0;i<n-1;i++) printf("%d ",c[i]);printf("%d\n",c[n-1]);}return 0;}
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