HDU5773：The All-purpose Zero

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.

给你一个数列 其中0可以变为任意一个数 问你最大递增子序列长度是多少

0可以变为任意数 把所有的0加入肯定是最优的 我们可以先找出除了0之外的最大递增子序列 再加上0的个数

要注意 1 2 0 3 这种情况会出现错误 我们需要让每个数减去在它之前的0的个数 这样我们就能得到严格递增的序列

前面的例子就变为1 2 0 2

再举个例子 12 0 3 5 0 0 0 6 9 变化之后为1 2 0 2 4 0 0 0 2 5 最大递增子序列为 1 2 4 5 最大长度为8

下面有用到lower_bound() 点击查看它的用法

#include<cstdio> #include<cstring> #include<algorithm> using namespace std; int a[100001]; int b[100001]; int d[100001]; int vis[100001]; int main() { int t,ans=0;; scanf("%d",&t); while(t--) { ans++; int n,m; scanf("%d",&n); int sum=0; int cnt=0; for(int i=1;i<=n;i++) { scanf("%d",&m); if(m==0) { sum++; } else { a[++cnt]=m-sum;//存入一个新的数组 } } if(cnt==0) { printf("Case #%d: %d\n",ans,sum); continue; } int len=1; d[1]=a[1]; for(int i=2;i<=cnt;i++) { if(a[i]>d[len]) { len++; d[len]=a[i]; } else { int pos=lower_bound(d+1,d+len,a[i])-d;//这里和下面的二分查找作用是一样的 用起来更方便一些 d[pos]=a[i]; /* int l=1; int r=len; int mid; int cut; while(l<=r) { mid =(l+r)/2; if(a[i]<d[mid]) { cut=mid; r=mid-1; } else { l=mid+1; } } d[cut]=a[i]; */ } } printf("Case #%d: %d\n",ans,len+sum); } return 0; }