hdu 5773 (The All-purpose Zero)

The All-purpose Zero

 

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.
 

 题意：求最长上升子序列，0可以变成任意数。

题解：LIS变形，一直没想到怎么处理中间的0，其实可以先去掉0，求出上升子序列，然后加上0的个数lower_bound(s,t,val),返回第一个大于等于val的数的位置，upper_bound(s,t,val)返回第一个大于val的数的位置，两者相减可以得到序列中k的个数。

#include<cstdio>#include<algorithm>#define INF 0x3f3f3f3f#define M 100100using namespace std;int s[M];int dp[M];int main(){int t,n,i,sum,ans;int cas=1;scanf ("%d",&t);while (t--){sum=0;ans=0;scanf ("%d",&n);for (i=0;i<n;i++){scanf ("%d",&s[i]);}fill(dp,dp+n,INF);for (i=0;i<n;i++){if (s[i]==0){ans++;}else*lower_bound(dp,dp+n,s[i]-ans)=s[i]-ans;//去除0}sum=lower_bound(dp,dp+n,INF)-dp;sort (s,s+n);sum+=upper_bound(s,s+n,0)-lower_bound(s,s+n,0);//可以直接sum+=ans;printf ("Case #%d: %d\n",cas++,sum);}return 0;}