POJ1157_LITTLE SHOP OF FLOWERS_动态规划

LITTLE SHOP OF FLOWERS

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 21143 Accepted: 9786

Description

You want to arrange the window of your flower shop in a most pleasant way. You have F bunches of flowers, each being of a different kind, and at least as many vases ordered in a row. The vases are glued onto the shelf and are numbered consecutively 1 through V, where V is the number of vases, from left to right so that the vase 1 is the leftmost, and the vase V is the rightmost vase. The bunches are moveable and are uniquely identified by integers between 1 and F. These id-numbers have a significance: They determine the required order of appearance of the flower bunches in the row of vases so that the bunch i must be in a vase to the left of the vase containing bunch j whenever i < j. Suppose, for example, you have bunch of azaleas (id-number=1), a bunch of begonias (id-number=2) and a bunch of carnations (id-number=3). Now, all the bunches must be put into the vases keeping their id-numbers in order. The bunch of azaleas must be in a vase to the left of begonias, and the bunch of begonias must be in a vase to the left of carnations. If there are more vases than bunches of flowers then the excess will be left empty. A vase can hold only one bunch of flowers. 

Each vase has a distinct characteristic (just like flowers do). Hence, putting a bunch of flowers in a vase results in a certain aesthetic value, expressed by an integer. The aesthetic values are presented in a table as shown below. Leaving a vase empty has an aesthetic value of 0. 
 

V A S E S

1

2

3

4

5

Bunches

1 (azaleas)

723-5-2416

2 (begonias)

521-41023

3 (carnations)

-21

5-4-2020
According to the table, azaleas, for example, would look great in vase 2, but they would look awful in vase 4. 

To achieve the most pleasant effect you have to maximize the sum of aesthetic values for the arrangement while keeping the required ordering of the flowers. If more than one arrangement has the maximal sum value, any one of them will be acceptable. You have to produce exactly one arrangement. 

Input

• The first line contains two numbers: F, V.
• The following F lines: Each of these lines contains V integers, so that A_ij is given as the j^th number on the (i+1)^st line of the input file.

• 1 <= F <= 100 where F is the number of the bunches of flowers. The bunches are numbered 1 through F. 
  
• F <= V <= 100 where V is the number of vases. 
  
• -50 <= Aij <= 50 where Aij is the aesthetic value obtained by putting the flower bunch i into the vase j.

Output

The first line will contain the sum of aesthetic values for your arrangement.

Sample Input

3 57 23 -5 -24 165 21 -4 10 23-21 5 -4 -20 20

Sample Output

53

大致题意：

花店有n种花和m种花盆。普通的花插在不同的花盆里能获得不同的美丽值。求最优插花策略使最终得到的美丽值最大化。插花的规则有两点：1)每个瓶子只能插一种花，且所有的花都要用上。 2)插花时花的顺序不能改变。如：第一种花所插的花瓶必须在第二种花所插花瓶的前面。

大体思路：

简单DP。

当 n == m 时，只有一种策略，自然就是最优解。

当 n < m 时，这个问题可以分解为两个子问题：1)把前n-1束花插进前m-1个瓶子里，然后把第n束花插进第m个瓶子里。2)把前当n束花插进前m个瓶子里。最优解就是两者中的最大值。

最终问题的最优解包含子问题的最优解。由此，以上子问题又可以以相同的方式继续划分。得到一个递归的式子： DP [ i , j ] = max { DP [ i - 1 , j-1 ]  + v [ i , j ] , D [ i , j - 1 ] } .

根据以上式子，自底向上地依次求解所有子问题的最优解，最终的到原问题的最优解。

#include<cstdio>#define max(x,y) x>y ? x : yint Map [105][105];int Dp  [105][105];int F, V;int main (){while(scanf("%d%d",&F,&V)!=EOF){for(int i=1; i<=F; i++)for(int j=1; j<=V; j++)scanf("%d", &Map[i][j]);for(int i=1; i<=F; i++)for(int j=i; j<=V-F+i; j++){if( i == j) Dp[i][j] = Dp[i-1][j-1] + Map[i][j];else Dp[i][j] = max(Dp[i-1][j-1]+Map[i][j], Dp[i][j-1]);}printf("%d\n",Dp[F][V]);}return 0;}