The All-purpose Zero

The All-purpose Zero

[b]Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 991    Accepted Submission(s): 475
[/b]

[align=left]Problem Description[/align]
?? 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.
 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
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.
 

[align=left]Sample Input[/align]

2
7
2 0 2 1 2 0 5
6
1 2 3 3 0 0

 

[align=left]Sample Output[/align]

Case #1: 5
Case #2: 5

HintIn the first case,you can change the second 0 to 3.So the longest increasing subsequence is 0 1 2 3 5.

<p>用STL做:</p><p>题解：求最长递增子序列，把0换成你需要的那个数，是他成为最长的递增子序列</p><p>函数lower_bound()在first和last中的<strong>前闭后开</strong>区间进行二分查找，返回大于或等于val的<strong>第一个元素</strong>位置。如果所有元素都小于val，则返回<strong>last</strong>的位置</p><p>举例如下：</p><p>一个数组number序列为：4,10,11,30,69,70,96,100.设要插入数字3,9,111.pos为要插入的位置的下标</p><p>则</p><p>pos = lower_bound( number, number + 8, 3) - number，pos = 0.即number数组的下标为0的位置。</p><p>pos = lower_bound( number, number + 8, 9) - number， pos = 1，即number数组的下标为1的位置（即10所在的位置）。
</p><p>pos = lower_bound( number, number + 8, 111) - number， pos = 8，即number数组的下标为8的位置（但下标上限为7，所以返回最后一个元素的下一个元素）。
</p><p>所以，要记住：函数lower_bound()在first和last中的<strong>前闭后开</strong>区间进行二分查找，返回大于或等于val的<strong>第一个元素</strong>位置。如果所有元素都小于val，则返回<strong>last</strong>的位置，且last的位置是越界的！！~</p>

#include<cstdio>
#include<cstring>
#include<cctype>
#include<cmath>
#include<set>
#include<map>
#include<list>
#include<queue>
#include<deque>
#include<stack>
#include<string>
#include<vector>
#include<iostream>
#include<algorithm>
#include<stdlib.h>
using namespace std;
typedef long long int LL;
const int INF=2e9+1e8;
const int MAXSIZE=1e6+10;
int dp[MAXSIZE];
int main()
{
int ncase,cases=0;
cin>>ncase;
while(ncase--)
{
int n,ans=-1;
scanf("%d",&n);
for(int i=0; i<=n+1; i++)
dp[i]=INF;
for(int i=0; i<n; i++)
{
int we,k;
scanf("%d",&we);
if(we)
{
k=lower_bound(dp,dp+n,we)-dp;
dp[k]=we;
ans=max(ans,k);
}
else
{
ans++;
for(int j=ans;j-1>=0;j--)
dp[j]=dp[j-1]+1;
dp[0]=0;
}
}
printf("Case #%d: %d\n",++cases,ans+1);
}
return 0;
}