多校4 The All-purpose Zero 5773

The All-purpose Zero

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1503    Accepted Submission(s): 717

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：给你一串序列，0可以变成任何数，问最长上升子序列

思路：nlogn 的LIS改编一下，遇到0把之前的i长度的子序列的最后一个值都改成i-1长度的子序列的最后一个值+1（为了让他尽量小）

代码：

#include<cstdio>#include<cstdlib>#include<algorithm>#include <iostream>using namespace std;int a[100010];int d[100010];int main(){    int t;    scanf("%d",&t);    int ca = t;    while(t--)    {        int n;        scanf("%d",&n);        for(int i=1;i<=n;++i)            scanf("%d",&a[i]);        int len = 1;        d[0] = -1;        for(int i = 1; i <= n; ++i)        {             if(a[i] == 0)             {                 for(int j = len;  j > 0; j--)                     d[j] = d[j-1] + 1;                 len++;             }             else             {                 if(a[i] > d[len - 1])                     d[len++] = a[i];                 else                     d[lower_bound(d + 1,d+len,a[i])-d] = a[i];             }        }        printf("Case #%d: %d\n",ca-t,len-1);    }    return 0;    }