HDU 5773 The All-purpose Zero （LIS变形）-优快云博客

本文链接：https://blog.youkuaiyun.com/Mr_Treeeee/article/details/72662707

探讨一种特殊场景下的最长递增子序列(LIS)问题解决方法，其中序列中的零元素可变为任意整数，旨在通过巧妙地利用零元素来最大化递增子序列的长度。

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The All-purpose Zero

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2078 Accepted Submission(s): 974

Problem Description

?? gets an sequence S with n intergers(0 < n <= 100000,0<= S[i] <= 1000000).?? has a magic so that he can change 0 to any interger(He does not need to change all 0 to the same interger).?? wants you to help him to find out the length of the longest increasing (strictly) subsequence he can get.

Input

The first line contains an interger T,denoting the number of the test cases.(T <= 10)
For each case,the first line contains an interger n,which is the length of the array s.
The next line contains n intergers separated by a single space, denote each number in S.

Output

For each test case, output one line containing “Case #x: y”(without quotes), where x is the test case number(starting from 1) and y is the length of the longest increasing subsequence he can get.

Sample Input

  
   2
7
2 0 2 1 2 0 5
6
1 2 3 3 0 0

Sample Output

  
   Case #1: 5
Case #2: 5


   
    
     Hint
    In the first case,you can change the second 0 to 3.So the longest increasing subsequence is 0 1 2 3 5.

Author

FZU

Source

2016 Multi-University Training Contest 4

题意：

0可以变成任何数，变化之后求LIS

point：

因为0是一定要用的。

变化原数组，把0移除，每个数前面有几个0就减几。因为两个数之间中间有0，0会补差。例如 1 0 2 ，原数组 2 这个数前面有个0，用了他之后，2-1>1成立才可以用。所以2这个位置至少要3才可以。

答案为数组LIS+zero的个数。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int a[100020];
int f[100020];
int n;
int main()
{
    int T;
    scanf("%d",&T);
    int p=0;
    while(T--)
    {
        int ans,cnt;
        ans=cnt=1;
        int num,zero;
        num=zero=0;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            int x;
            scanf("%d",&x);
            if(x==0)
            {
                zero++;
            }
            else
            {
                a[++num]=x-zero;
            }
        }
        memset(f,0,sizeof f);
        f[1]=a[1];
        for(int i=2;i<=num;i++)
        {
            if(a[i]>f[cnt]) f[++cnt]=a[i];
            else
            {
                f[lower_bound(f+1,f+1+cnt,a[i])-f]=a[i];
            }
        }
        if(zero==n) cnt=0;//cnt=1会影响答案。
         printf("Case #%d: %d\n",++p,cnt+zero);
    }

}