HDU 5773The All-purpose Zero

The All-purpose Zero

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 231 Accepted Submission(s): 101

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 0 2 1 2 0

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.

0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。

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。

/* ***********************************************
Author        :guanjun
Created Time  :2016/7/28 15:51:43
File Name     :p410.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 100010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
    int x,y;
};
struct cmp{
    bool operator()(Node a,Node b){
        if(a.x==b.x) return a.y> b.y;
        return a.x>b.x;
    }
};

bool cmp(int a,int b){
    return a>b;
}
int d[maxn];
int a[maxn];
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int T,n,cnt;
    cin>>T;
    for(int t=1;t<=T;t++){
        cin>>n;
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        int len=0;
        cnt=0;
        for(int i=1;i<=n;i++){
            if(a[i]==0)cnt++;
            else {
                a[i]-=cnt;
                int pos=lower_bound(d,d+len,a[i])-d;
                if(pos==len){
                    d[len++]=a[i];
                }
                else d[pos]=a[i];
            }
        }
        printf("Case #%d: %d
",t,len+cnt);
    }
    return 0;
}