hdoj 5773 <变相LIS>

The All-purpose Zero

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1062    Accepted Submission(s): 511

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

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​​

正在品味这种方法----好聪明--智商压制我-.-

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
int shu[101000];
int main()
{
  int t,n;scanf("%d",&t);
  for (int ca=1;ca<=t;ca++)
  {
    scanf("%d",&n);
    int a,jk=0,kp=0;
    while (n--)
    {
      scanf("%d",&a);
      if (a)
      {
        a-=jk;
        if (kp==0)
        shu[kp++]=a;
        else
        {
          if (a>shu[kp-1])
          shu[kp++]=a;
          else
          {
            int ll=lower_bound(shu,shu+kp,a)-shu;
            shu[ll]=a;
          }
        }
      }
      else
      jk++;
    }
    printf("Case #%d: %d\n",ca,jk+kp);
  }
  return 0;
}