hdoj 5500 Reorder the Books
Reorder the Books
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 919 Accepted Submission(s):
501
Problem Description
dxy has a collection of a series of books called "The
Stories of SDOI",There are n(n≤19)
books in this series.Every book has a number from 1
to n
.
dxy puts these books in a book stack with the order of their numbers increasing from top to bottom. dxy takes great care of these books and no one is allowed to touch them.
One day Evensgn visited dxy's home, because dxy was dating with his girlfriend, dxy let Evensgn stay at home himself. Evensgn was curious about this series of books.So he took a look at them. He found out there was a story about "Little E&Little Q". While losing himself in the story,he disrupted the order of the books.
Knowing that dxy would be back soon,Evensgn needed to get the books ordered again.But because the books were too heavy.The only thing Evensgn could do was to take out a book from the book stack and and put it at the stack top.
Give you the order of the disordered books.Could you calculate the minimum steps Evensgn would use to reorder the books? If you could solve the problem for him,he will give you a signed book "The Stories of SDOI 9: The Story of Little E" as a gift.
dxy puts these books in a book stack with the order of their numbers increasing from top to bottom. dxy takes great care of these books and no one is allowed to touch them.
One day Evensgn visited dxy's home, because dxy was dating with his girlfriend, dxy let Evensgn stay at home himself. Evensgn was curious about this series of books.So he took a look at them. He found out there was a story about "Little E&Little Q". While losing himself in the story,he disrupted the order of the books.
Knowing that dxy would be back soon,Evensgn needed to get the books ordered again.But because the books were too heavy.The only thing Evensgn could do was to take out a book from the book stack and and put it at the stack top.
Give you the order of the disordered books.Could you calculate the minimum steps Evensgn would use to reorder the books? If you could solve the problem for him,he will give you a signed book "The Stories of SDOI 9: The Story of Little E" as a gift.
Input
There are several testcases.
There is an positive integer T(T≤30) in the first line standing for the number of testcases.
For each testcase, there is an positive integer n in the first line standing for the number of books in this series.
Followed n positive integers separated by space standing for the order of the disordered books,the ith integer stands for the ith book's number(from top to bottom).
Hint:
For the first testcase:Moving in the order of book3,book2,book1 ,(4,1,2,3)→(3,4,1,2)→(2,3,4,1)→(1,2,3,4) ,and this is the best way to reorder the books.
For the second testcase:It's already ordered so there is no operation needed.
There is an positive integer T(T≤30) in the first line standing for the number of testcases.
For each testcase, there is an positive integer n in the first line standing for the number of books in this series.
Followed n positive integers separated by space standing for the order of the disordered books,the ith integer stands for the ith book's number(from top to bottom).
Hint:
For the first testcase:Moving in the order of book3,book2,book1 ,(4,1,2,3)→(3,4,1,2)→(2,3,4,1)→(1,2,3,4) ,and this is the best way to reorder the books.
For the second testcase:It's already ordered so there is no operation needed.
Output
For each testcase,output one line for an integer
standing for the minimum steps Evensgn would use to reorder the books.
Sample Input
2
4
4 1 2 3
5
1 2 3 4 5
Sample Output
3
0
题意:一摞书顺序乱了,让你按照序号从小到大的顺序将其排列整齐,一次只能拿一本书,且必须放在最上边,问最少多少步可以实现
题解:先算出已经按顺序排好的书本的个数sum(注意这个序列并不一定连续,而且必须是从某个数到n的序列),用n减去sum即可
#include<stdio.h> #include<string.h> #include<algorithm> #define MAX 100100 using namespace std; int a[MAX]; int main() { int n,m,t,i,j; scanf("%d",&t); int maxx,ans; while(t--) { scanf("%d",&n); maxx=0; for(i=1;i<=n;i++) { scanf("%d",&a[i]); if(maxx<a[i]) { maxx=a[i]; ans=i; } } int ant=ans,sum=0; for(j=ans;j>=1;j--) { if(a[j]==a[ant]-1) { sum++; ant=j; } } printf("%d\n",n-sum-1); } return 0; }