POJ 1157解题报告

LITTLE SHOP OF FLOWERS

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 18196 Accepted: 8381

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 thatA_ij is given as thej^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

Source

IOI 1999

        这道题是在最近学习dp的时候看到的。由于刚刚接触dp，对dp还不是特别理解。不过在本题中dp的状态转移方程还是比较好写的。本题中自我感觉比较坑的是不放花的花瓶对应的a[i][j]=0,我在写状态转移方程时专门判断了下不为0的情况，但现在想想这样是不对的。还有本题的初始化和最终的求最优解也是比较容易错的，因为有正负数存在，所以初始化的时候应该将所有的dp[i][j]初始化为-INF(INF为无穷大的数)，然后在写状态转移方程时最好不要写成dp[i][j]=max(dp[i][j],dp[i][k]+a[i][j])这种形式，因为这样如果dp[i][k]本身就是-INF,相当于达不到，这样转移是没有意义的。因此状态转移方程写成if语句判断的形式会比较好。WA了好几发，感觉就是自己一开始在这里写残了。后来改完就对了。初学dp，还是不太熟练。。加油！
附上AC的代码，虽然不是很快，但毕竟进步一点点。

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>#define maxn 1000100#define CLR(x) memset(x,0,sizeof(x))using namespace std;const int INF=(1<<30);int a[200][200],dp[200][200];int main(){    int F,V;    scanf("%d%d",&F,&V);    for(int i=0; i<=F; i++)        for(int j=0; j<=V; j++)            dp[i][j]=-INF;    for(int i=1; i<=F; i++)        for(int j=1; j<=V; j++)            scanf("%d",&a[i][j]);    for(int i=1; i<=F; i++)    {        for(int j=1; j<=V; j++)        {            if(i==1)            {                dp[i][j]=a[i][j];                continue;            }            for(int k=1; k<j; k++)            {                if(dp[i-1][k]!=-INF&&dp[i-1][k]+a[i][j]>dp[i][j])                    dp[i][j]=dp[i-1][k]+a[i][j];            }        }    }    int res=dp[F][1];    for(int i=1; i<=V; i++)    {        if(dp[F][i]>res)            res=dp[F][i];    }    printf("%d\n",res);}