LITTLE SHOP OF FLOWERS

Time Limit: 1000MS Memory Limit: 10000K

Total Submissions: 19877 Accepted: 9153

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)

7 23 -5 -24 16

2 (begonias)

5 21 -4 10 23

3 (carnations)

-21

5 -4 -20 20

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 Aij is given as the jth 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 5

7 23 -5 -24 16

5 21 -4 10 23

-21 5 -4 -20 20

Sample Output

53

、

题意：就是给你一个二维数组，每一行只能选一个，并且选择的数字从上倒下，必须是从左到右的顺序，求最大的和

。动态规划的题目

状态转移方程 dp[i][j]=max(dp[i-1][j-1]+a[i][j],dp[i][j-1]);

这个题目需要注意的是当dp[i][i]即i=j的时候状态转移方程是

dp[i][i]=dp[i-1][i-1]+a[i][i];

为什么呢？因为当i=j的时候，之前的几列都可以确定了，因为要求顺序是从左到右啊，

还有这道题目让我对动态规划更深层的认识是：

时时刻刻要记住动态规划是从子问题不断递推而来，解决最终问题的，一开始我不明白为什么 dp[i][j]=max(dp[i-1][j-1]+a[i][j],dp[i][j-1]);完全没有考虑dp[i-1][j….m],就是对dp[i][j]来说，i-1行，大于j也可以取到啊，为什么不考虑呢？原来，这是因为，对于dp[i][j]这个子问题来说，整个二维数组最大的就是行长度就是j无需考虑比j大的

#include <iostream>

#include <string.h>

#include <math.h>

#include <algorithm>

#include <stdlib.h>

using namespace std;

int dp[100][100];//i行，j列，最大的值

int f,v;

int a[100][100];

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",&a[i][j]);

            }

        }

        memset(dp,0,sizeof(dp));

        for(int i=1;i<=f;i++)

        {

             dp[i][i]=dp[i-1][i-1]+a[i][i];

            for(int j=i+1;j<=v;j++)

            {

               dp[i][j]=max(dp[i-1][j-1]+a[i][j],dp[i][j-1]);

            }

        }

        printf("%d\n",dp[f][v]);

    }

    return 0;

}