HDU：5773 The All-purpose Zero（LIS-n*logn解法+思维+技巧）

The All-purpose Zero

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1618    Accepted Submission(s): 774

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.
 

 

Author

FZU

 

Source

2016 Multi-University Training Contest 4

 

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题目大意：给了你一个序列，其中0可以修改成任意数字（包括负数），问修改之后最长子序列长度为多少。

解题思路：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。（注意全0的时候特判下，这个方法不符合）

代码如下：

 

#include <cstdio>int b[100010];int d[100010];int ans;int erfen(int x){int l=1,r=ans;int mid,pos;while(l<=r){mid=(l+r)/2;if(d[mid]>x){pos=mid;r=mid-1;}else{l=mid+1;}}return pos;}int main(){int t;scanf("%d",&t);int k=1;while(t--){int n;scanf("%d",&n);int num=0;//0的数量 int hao=0;//新数组角标 for(int i=1;i<=n;i++)//输入数据的同时构建新数组 {int x;scanf("%d",&x);if(x==0){num++;}else{hao++;b[hao]=x-num;}}if(hao==0)//全0的时候特判一下 {printf("Case #%d: ",k++);printf("%d\n",n);continue;}d[1]=b[1];ans=1;//记录LIS长度 for(int i=2;i<=hao;i++){if(b[i]>d[ans]){ans++;d[ans]=b[i];}else{d[erfen(b[i])]=b[i];}}printf("Case #%d: ",k++);printf("%d\n",ans+num);}return 0;}