[HDU - 5773] The All-purpose Zero（动态规划&&LIS）-优快云博客

本文介绍了一种解决最长递增子序列问题的算法，特别考虑了序列中0可以转换为任意整数的情况。通过记录0的数量并在计算过程中进行调整，最终得出最长递增子序列的长度。

Link：http://acm.hdu.edu.cn/showproblem.php?pid=5773

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.

题解：

题意就是0可以变成任何值求LIS

步骤：

b[i]==0时记下i之前0的个数, 记为k
b[i]!=0时存在新数组a里，a[j++]=b[i]-k；
ans = 新数组a的LIS+k

对于第2点，例如当数列为 1 0 0 0 2 时，LIS= 4（1 2 3 4 2），1、2 之间没有整数，2减去它前面0的个数为-1, （1 -1）的LIS=1，再加3可得LIS=4

Code：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<queue>
#include<stack>
#define INF 0x3f3f3f
using namespace std;
const int maxn=1e6+10;
int a[maxn];
int dp[maxn];
int b[maxn];
int main()
{
    int t,n,k,c=1;
    scanf("%d",&t);
    while(t--)
    {
        k=0;
        int j=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&b[i]);
            dp[i]=INF;
            if(b[i]==0)
            {
                k++;
            }

            else
            {
                b[i]-=k;
                a[j++]=b[i];
            }   
        }
        for(int i=0;i<j;i++)
            *lower_bound(dp,dp+n,a[i])=a[i];
        printf("Case #%d: %d\n",c++,lower_bound(dp,dp+n,INF)-dp+k);
    }
return 0;
}