The All-purpose Zero

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1097 Accepted Submission(s): 271

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

Case #1: 5 Case #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：有一个序列，含有n个数，序列中的0可以变成任意数，问你变换完后最长上升子序列的长度是多少

显然在最后面的0可以作为最长上升子序列的元素，那么前面的0怎么变换呢，思路是这样的：把0去掉，做LIS 为了保证是添上0变换的数字后是上升的，把每个非0数减去前面包含的0的个数，然后len加上0的个数即可

样例一 2 0 2 1 2 0 5 -> 2 (0) 1 0 1 (0) 3 新序列是 2 1 0 1 3 去掉了两个0，新序列LIS=3

那么原序列LIS=3+2 =5；

#include <stdio.h>
#include <iostream>
#include <stdlib.h>using namespace std;
const int inf=0x3f3f3f3f,maxn=1e5+10;
int a[maxn],c[maxn],len;
int find(int R,int x)
{int L=1,mid;while(L<=R){mid=(L+R)/2;if(c[mid]==x)return mid;else if(c[mid]<=x)L=mid+1;else R=mid-1;}return L;
}int main()
{int n,m,T=1,t;int sum,j,zero,tot;scanf("%d",&t);while(t--){scanf("%d",&n);sum=0;tot=0;int x;for(int i=1;i<=n;i++){scanf("%d",&x);if(x==0)sum++;elsea[tot++]=x-sum;}if(tot==0){printf("Case #%d: %d\n",T++,n);continue;}len=0;c[0]=-inf;for(int i=0;i<tot;i++){if(a[i]>c[len])j=++len;elsej=find(len,a[i]);c[j]=a[i];}/*for(int i=1;i<=len;i++)cout<<c[i]<<" ";cout<<endl;*/printf("Case #%d: %d\n",T++,len+sum);}
}

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The All-purpose Zero

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