The All-purpose Zero

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 279 Accepted Submission(s): 129

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.

Author
FZU

Source
2016 Multi-University Training Contest 4

題意：
給你一個數組，其中的0可以變成任何數，求最長的LIS是多長。

題解：
因為0可以變成任何數，那麼所有的0是絕對要用上的（就算強行用上，頂多是使原LIS保持長度，不會出現用0減長度的情況）。所以把所有的0單獨拿出來後，看剩下數的LIS，為了求出的LIS保證在原位置加入0後仍然是LIS，我們將原本非0的元素都減去前方0的個數再去求LIS，這樣就能保證如果此數在LIS中，那麼加入0以後也一定是LIS。
因為資料大小和時限的原因，用一個二分nlogn的LIS即可。

//
//  main.cpp
//  160728-4
//
//  Created by 袁子涵 on 16/7/28.
//  Copyright © 2016年 袁子涵. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>

using namespace std;
const int MAXN=100005;
int t,cas;
long long int n,b[MAXN],out,tmp;
long long int num[MAXN];
long long int Search(long long int num,long long int low,long long int high)
{
    long long int mid;
    while(low<=high)
    {
        mid=(low+high)/2;
        if(num>=b[mid])  low=mid+1;
        else   high=mid-1;
    }
    return low;
}
long long int DP(long long int n)
{
    long long int i,len,pos;
    b[1]=num[1];
    len=1;
    for(i=2;i<=n;i++)
    {
        if(num[i]>b[len])
        {
            len=len+1;
            b[len]=num[i];
        }
        else
        {
            pos=Search(num[i],1,len);
            b[pos]=num[i];
        }
    }
    return len;
}
long long int total=0;
bool flag;
int main(int argc, const char * argv[]) {
    scanf("%d",&t);
    while (t--) {
        long long int now=0;
        total=0;
        cas++;
        scanf("%lld",&n);
        for (long long int i=1; i<=n; i++) {
            scanf("%lld",&tmp);
            if (tmp==0)
                total++;
            else
                num[++now]=tmp-total;
        }
        out=DP(now);
        if (total==n)
            out=0;
        cout << "Case #" << cas << ": " << out+total << endl;
    }
    return 0;
}