Bellovin HDU - 5748 （最长上升子序列）

Bellovin

HDU - 5748

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 sequence b1,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 sequence a1,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

.

Input There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first contains an integer n (1≤n≤100000) -- the length of the sequence. The second line contains n integers a1,a2,...,an (1≤ai≤109). Output For each test case, output n integers b1,b2,...,bn (1≤bi≤109)denoting the lexicographically smallest sequence.
Sample Input

3
1
10
5
5 4 3 2 1
3
1 3 5

Sample Output

1
1 1 1 1 1
1 2 3

题意：给n个数字表示a序列，求字典序最小的b序列中每个位置的LIS跟a序列一样

思路：字典序最小其实就是a序列的LIS

 其思想源自最长上升子系列的lower_bound的优化   最长上升子序列问题l

#include 
#include 
#include 
#include 
#define INF 0x3f3f3f3f
using namespace std;

int a[100100];
int dp[100100];//dp[i]储存长度为i的上升子序列中末尾元素最小值
int ans[100100];
int n;
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        int i;
        for(i = 1; i <= n; i++){
            scanf("%d",&a[i]);
            dp[i] = INF;
        }
        int pos;
        //这道题不是让求最长上升子序列，而是求出以每个元素为末尾时的最长上升序列
        for(i = 1; i <= n; i++){
            pos = lower_bound(dp+1,dp+1+n,a[i])-dp;//每次查找
            ans[i] = pos;//记录位置
            dp[pos] = a[i];//更新dp数组
        }
        for(i = 1; i <= n; i++){
            if(i==1)
                printf("%d",ans[i]);
            else
                printf(" %d",ans[i]);
        }
        printf("\n");
    }
    return 0;
}