The All-purpose Zero

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

272 0 2 1 2 0 561 2 3 3 0 0

Sample Output

Case #1: 5Case #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.         

官方题解：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。

• 如果0位于最长上升子序列两边，这两个零要加进去是显然的
• 举个例子：1 2 3 0 4 5 6->预处理后：1 2 3 0 3 4 5，最长是1 2 3 0 4（5） 5（6），括号里是原先的数，这个例子里0是替代了原序列0右边的4

• 有许多个0夹于LIS的，同理可得

#include<cstdio>  #include<cstring>  #include<algorithm>  using namespace std;  int a[100009];int main(){int t,n,i,x,h=1,pos;scanf("%d",&t);while(t--){int p=0,q=0;scanf("%d",&n);for(i=0;i<n;i++){scanf("%d",&x);if(x){x=x-p;if(q==0){a[q++]=x;}else{if(x>a[q-1])a[q++]=x;else{pos=lower_bound(a,a+q,x)-a;//注意一下，这里是q，不是n,可以看成sort快排那样，是数组的范围。a[pos]=x;}}}elsep++;}printf("Case #%d: ",h++);printf("%d\n",p+q);}return 0;}