HDU 5773 The All-purpose Zero【LIS变形】

The All-purpose Zero

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

Total Submission(s): 2317    Accepted Submission(s): 1053

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

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.

首先一点：我们把所有0都用上是肯定不亏的。

接下来：我们把剩下的非0数求LIS， 但是， 这些0有可能不能全部插入LIS中， 那么势必要在LIS中剔出一些数。 一个很简单的方法， 我们把每一个数减去他前面的0的个数， 这样， 相当于先为0留出了空间， 因为是全都减小， 相对大小是不变的。

#include <cstdio>  #include <algorithm>  #define INF 0x3f3f3f  using namespace std;  int dp[100010],a[100010];  int main()  {      int n,i,j,T;      scanf("%d",&T);    int Case=1;    while(T--)      {      scanf("%d",&n);        for(i=0;i<n;++i)          {              scanf("%d",&a[i]);              dp[i]=INF;          }int ant=0;          for(i=0;i<n;++i)           {           if(a[i]==0)         {         ant++;         continue;}a[i]-=ant; *lower_bound(dp,dp+n,a[i])=a[i];  }          printf("Case #%d: %d\n",Case++,lower_bound(dp,dp+n,INF)-dp+ant);       }      return 0;  }