Hdu-5773 The All-purpose Zero(LIS变形)

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：给你一个含零的序列，问其最长不降子序列（严格单调），其中零可以化为任何数。分析：如果在最优方案中的两个数a b 中间存在一个0，那么用0代替a 或 b一定不会使结果变差，所以我们要选上所有的0，于是我们把每个非零数减去它前边零的个数后求LIS。
#include <cstdio>#include <stack>#include <queue>    #include <vector>    #include <cstdio>    #include <utility>    #include <cstring>    #include <iostream>    #include <algorithm>    #define INF 0x3f3f3f3fusing namespace std;int T,n,tot,num,x,a[100005],f[100005];int main(){scanf("%d",&T);for(int t = 1;t <= T;t++){tot = num = 0;scanf("%d",&n);for(int i = 1;i <= n;i++){scanf("%d",&x);if(x) {x -= tot;a[++num] = x;}else tot++;}memset(f,INF,sizeof(f));for(int i = 1;i <= num;i++){int *pos = lower_bound(f+1,f+2+num,a[i]);*pos = a[i]; }f[0] = 0;for(int i = num;i >= 0;i--) if(f[i] < f[100004]) {  printf("Case #%d: %d\n",t,i+tot);  break; }}}