SGU104 Little shop of flowers

104. Little shop of flowers

time limit per test: 0.5 sec. 
memory limit per test: 4096 KB

 PROBLEM

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.

ASSUMPTIONS

• 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 £ A_ij £ 50 where A_ij is the aesthetic value obtained by putting the flower bunch i into the vase j.

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.

 Output

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

• The second line must present the arrangement as a list of F numbers, so that the k’th number on this line identifies the vase in which the bunch k is put.

 Sample Input

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

Sample Outpu

53 

2 4 5

题目大意:假设你想以最美观的方式布置花店的橱窗。你有F束花，每束花的品种都不一样，同时，你至少有同样数量的花瓶，被按顺序摆成一行。花瓶的位置是固定的，并从左至右，从1至V 顺序编号，V 是花瓶的数目，编号为1 的花瓶在最左边，编号为V 的花瓶在最右边。花束则可以移动，并且每束花用1 至F 的整数唯一标识。标识花束的整数决定了花束在花瓶中排列的顺序，即如果i＜j，则花束i 必须放在花束j 左边的花瓶中。每束花插入不同的花瓶会产生相应的美观值，为取得最佳美学效果，你必须在保持花束顺序的前提下，使花束的摆放取得最大的美学值。如果具有最大美学值的摆放方式不止一种，则其中任何一种摆放方式都可以接受，但你只右输出其中一种摆放方式。

题解：简单的动态规划。自己YY了个O(n^3)的方程。。。不过目测N=100会挂掉。。。但又没想到什么优化方法，无奈只好去看题解（动态规划的题目做得太少了，真心需要多做题才会啊），方程是这样的f[i][j]=max(f[i][j-1],f[i-1][j-1]+a[i][j])。方程应该很好理解。。。可我就是二逼得没想到啊！！！！

#include<stdio.h>
#include<string.h>
#define MAXN 105
long g[MAXN][MAXN],path[MAXN];
int a[MAXN][MAXN],visit[MAXN][MAXN];
int f,v;
int main(void)
{
    long i,j;
    scanf("%d%d",&f,&v);
    for(i=1; i<=f; i++)
        for(j=1; j<=v; j++)
            scanf("%d",&a[i][j]);
    memset(g,0,sizeof(g));
    memset(visit,0,sizeof(visit));
    for(i=1; i<=f; i++)
    {
        g[i][i]=g[i-1][i-1]+a[i][i];
        visit[i][i]=1;
        for(j=i+1; j<=v-(f-i); j++)
            if (g[i][j-1]<g[i-1][j-1]+a[i][j])
            {
                g[i][j]=g[i-1][j-1]+a[i][j];
                visit[i][j]=1;
            }
            else
                g[i][j]=g[i][j-1];
    }

    printf("%ld\n",g[f][v]);
    i=f;
    for(j=v; j>=1; j--)
        if(visit[i][j])
            path[i--]=j;
    for(i=1; i<f; i++)
        printf("%ld ",path[i]);
    printf("%ld\n",path[i]);
    return 0;
}