題意   你有f束花和v個花瓶  每束花放在不同的花瓶中都會有不同的價值  並且花的相對順序不能改變  也就是說第i束花放在第j個花瓶中  那麼第i+1朵花要放在j以後的花瓶中  

令d[i][j]表示第i朵花放在第[j]個瓶子中前i朵花的最大價值  有狀態轉移方程 d[i][j]=max{d[i-1][1~j-1]}+val[i][j];

題目有幾個坑點   可能所有價值都為負數  必須所有花都放完

#include<cstdio>  
#include<cstring>  
using namespace std;  
#define M 205  
int val[M][M],d[M][M],pre[M][M],f,v,key;  
  
void print(int i,int j)  
{  
    if(pre[i][j])  
    {  
        print(i-1,pre[i][j]);  
        printf(" %d",j);  
    }  
    else  
        printf("%d",j);  
}  
  
int main()  
{  
    scanf("%d%d",&f,&v);  
    memset(d,0x8f,sizeof(d));  
  
    for(int i=1; i<=f; ++i)  
        for(int j=1; j<=v; ++j)  
            scanf("%d",&val[i][j]);  
  
    int ans=d[0][0];  
    //注意要把ans初始化為負無窮 可能所有的val都為負；這裡wa了好久；  
    d[0][0]=0;  
    //0朵花價值當然為0了；  
      
    for(int i=1; i<=f; ++i)  
        for(int j=1; j<=v; ++j)  
        {  
            for(int k=0; k<j; ++k)  
                if(d[i-1][k]+val[i][j]>d[i][j])  
                {  
                    d[i][j]=d[i-1][k]+val[i][j];  
                    pre[i][j]=k;  
                }  
        }  
    for(int j=1; j<=v; ++j)  
        if(d[f][j]>ans)  
        {  
            ans=d[f][j];  
            key=j;  
        }  
    //這裡要在最後一層更新ans 因為要保證所有花都放進去；  
  
    printf("%d\n",ans);  
    print(f,key);  
    printf("\n");  
  
    return 0;  
}  
 

Little shop of flowers

PROBLEM

You want to arrange the window of your flower shop in a most pleasant way. You haveFbunches 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 throughV, whereVis the number of vases, from left to right so that the vase 1 is the leftmost, and the vaseV

is the rightmost vase. The bunches are moveable and are uniquely identified by integers between 1 andF. 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 bunchimust be in a vase to the left of the vase containing bunchjwheneveri<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.

ASSUMPTIONS

1 ≤F≤ 100 whereFis the number of the bunches of flowers. The bunches are numbered 1 throughF.F≤V≤ 100 whereVis the number of vases.-50£A_ij£50 whereA_ijis the aesthetic value obtained by putting the flower bunchiinto the vasej.

Input

The first line contains two numbers:F,V.The followingFlines: Each of these lines containsVintegers, so thatA_ijis given as thej’th number on the (i+1)’st line of the input file.

Output

The first line will contain the sum of aesthetic values for your arrangement.The second line must present the arrangement as a list ofFnumbers, so that thek’th number on this line identifies the vase in which the bunchkis put.

Sample Input

3 5 
7 23 -5 -24 16
5 21 -4 10 23
-21 5 -4 -20 20

Sample Output

53 
2 4 5