HDU-5773-The All-purpose Zero

题目链接：点击打开链接

The All-purpose Zero

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1567    Accepted Submission(s): 750

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.
   
  

大意：在一个序列里面求最长严格上升子序列，其中0可以任意变换为其他数。

我这个智障也是看了别人的思路，才写了点题解。原文在这：点击打开链接

题解： 0 可以转化成任意整数，包括负数，显然求LIS时尽量把0都放进去必定是正确的。因此我们可以把0拿出来，对剩下的做 O( nlogn ) 的 LIS，统计结果的时候再算上 0 的数量。为了保证严格递增，我们可以将每个权值 S [ i ]减去 i 前面 0 的个数，再做 LIS，就能保证结果是严格递增的。

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int MAX=1e5+10;
int n;
int dp[MAX],a[MAX];
int main()
{
	int t,text=0;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		int x,cnt=0,num=0;
		for(int i=0;i<n;i++)
		{
			scanf("%d",&x);
			if(x==0)	cnt++;
			else	a[num++]=x-cnt;
		}
		if(num==0)
		{
			printf("Case #%d: %d\n",++text,n);
			continue;
		}
		int k=0;
		memset(dp,0,sizeof(dp));
		dp[0]=a[0];
		for(int i=1;i<num;i++)
		{
			if(a[i]>dp[k])
				dp[++k]=a[i];
			else
			{
				int pos=lower_bound(dp,dp+k,a[i])-dp;
				dp[pos]=a[i];
			}
		}
		printf("Case #%d: %d\n",++text,k+1+cnt);
	}
	return 0;
}