HDOJ-----5748Bellovin(LIS)

最新推荐文章于 2019-12-23 19:07:07 发布

pupil_blue

最新推荐文章于 2019-12-23 19:07:07 发布

阅读量229

点赞数

CC 4.0 BY-SA版权

分类专栏： LIS/LCS

本文链接：https://blog.youkuaiyun.com/blue_hpu/article/details/52197828

LIS/LCS 专栏收录该内容

7 篇文章

订阅专栏

本文介绍了一个算法问题，目标是找到一个序列，使得该序列与给定序列在每个位置上的最长递增子序列长度相同，且该序列在字典序上是最小的。文章通过示例解释了如何求解这一问题，并提供了一段C++代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Bellovin

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1100 Accepted Submission(s): 496

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 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

(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

Sample Output

1
1 1 1 1 1
1 2 3

寻找与原序列每个元素的最长递增子序列相等的序列的最小值

例如 5 4 3 2 1

每个元素的最长递增子序列都是1，所以序列中每个元素的LIS都是1的序列最小值为1 1 1 1 1（严格递增）

即求出原序列中以每个元素结尾的LIS长度

#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 100010
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
int dp[maxn], cnt[maxn], vis[maxn];
int ans, n;
void lis(){
    int flag;
    ans = 0;
    for(int i = 1; i <= n; i++){
        if(dp[ans] < cnt[i]){
            flag = ++ans;
        }
        else{
            flag = lower_bound(dp+1, dp+ans+1, cnt[i])-dp;
        }
        dp[flag] = cnt[i];
        vis[i] = flag;
    }
}
int main(){
	int t, num;
	scanf("%d", &t);
	while(t--){
        memset(dp, 0, sizeof(dp));
        scanf("%d", &n);
        for(int i = 1; i <= n; i++){
            scanf("%d", &cnt[i]);
        }
        lis();
        for(int i = 1; i <= n; i++){
            printf("%d%c", vis[i], i == n ? '\n' : ' ');
        }
	}
	return 0;
}