題目

Source

http://www.lightoj.com/volume_showproblem.php?problem=1283

Description

You are a librarian. You keep the books in a well organized form such that it becomes simpler for you to find a book and even they look better in the shelves.

One day you get n new books from one of the library sponsors. And unfortunately they are coming to visit the library, and of course they want to see their books in the shelves. So, you don't have enough time to shelve them all in the shelf in an organized manner since the heights of the books may not be same. But it's the question of your reputation, that's why you have planned to shelve them using the following idea:

1) You will take one book from the n books from left.
2) You have exactly one shelf to organize these books, so you may either put this book in the left side of the shelf, right side of the shelf or you may not put it in the shelf. There can already be books in the left or right side. In such case, you put the book with that book, but you don't move the book you put previously.
3) Your target is to put the books in the shelf such that from left to right they are sorted in non-descending order.
4) Your target is to put as many books in the shelf as you can.

You can assume that the shelf is wide enough to contain all the books. For example, you have 5 books and the heights of the books are 3 9 1 5 8 (from left). In the shelf you can put at most 4 books. You can shelve 3 5 8 9, because at first you got the book with height 3, you stored it in the left side of the shelf, then you got 9 and you put it in the right side of the shelf, then you got 1 and you ignored it, you got 5 you put it in the left with 3. Then you got 5 and you put it in left or right. You can also shelve 1 5 8 9 maintaining the restrictions.

Now given the heights of the books, your task is to find the maximum number of books you can shelve maintaining the above restrictions.

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 100). Next line contains n space separated integers from [1, 105]. The ith integer denotes the height of the ith book from left.

Output

For each case, print the case number and the maximum number of books that can be shelved.

Sample Input

2
5
3 9 1 5 8
8
121 710 312 611 599 400 689 611

Sample Output

Case 1: 4
Case 2: 6

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分析

題目大概說有n本書，要依次把它們放到書架，可以放到書架的左邊或者右邊挨著已經放好的書的下一個位置，當然也可以選擇不放。放好后要保證書的高度從左到右非遞減。問最多能放上幾本書。

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n才100，果斷這么表示狀態：

• dp[i][j][k]表示放置前i本書，書架的左邊最后面的書是第j本且書架右邊最前面的書是第k本，最多能放的書數

轉移我用我為人人，通過dp[i-1]的狀態選擇將第i本書放到左邊還是右邊或者不放來轉移并更新dp[i]的狀態值。

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代碼

#include<cstdio> #include<cstring> #include<algorithm> using namespace std; int d[111][111][111]; int main(){int t,n,a[111];scanf("%d",&t);for(int cse=1; cse<=t; ++cse){scanf("%d",&n);for(int i=1; i<=n; ++i){scanf("%d",&a[i]);}memset(d,-1,sizeof(d));d[0][0][0]=0;for(int i=0; i<n; ++i){for(int j=0; j<=i; ++j){for(int k=0; k<=i; ++k){if(d[i][j][k]==-1) continue;d[i+1][j][k]=max(d[i+1][j][k],d[i][j][k]);if(j==0 && k==0){d[i+1][i+1][0]=1;d[i+1][0][i+1]=1;}else if(j==0){if(a[k]>=a[i+1]) d[i+1][i+1][k]=max(d[i+1][i+1][k],d[i][j][k]+1);if(a[k]>=a[i+1]) d[i+1][j][i+1]=max(d[i+1][j][i+1],d[i][j][k]+1);}else if(k==0){if(a[i+1]>=a[j]) d[i+1][i+1][k]=max(d[i+1][i+1][k],d[i][j][k]+1);if(a[i+1]>=a[j]) d[i+1][j][i+1]=max(d[i+1][j][i+1],d[i][j][k]+1);}else{if(a[j]<=a[i+1] && a[i+1]<=a[k]){d[i+1][i+1][k]=max(d[i+1][i+1][k],d[i][j][k]+1);d[i+1][j][i+1]=max(d[i+1][j][i+1],d[i][j][k]+1);}}}}}int res=0;for(int i=0; i<=n; ++i){for(int j=0; j<=n; ++j){res=max(res,d[n][i][j]);}}printf("Case %d: %d\n",cse,res);}return 0; }

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轉載于:https://www.cnblogs.com/WABoss/p/5765310.html

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