POJ1157——LITTLE SHOP OF FLOWERS

LITTLE SHOP OF FLOWERS
Time Limit: 1000MS
 
Memory Limit: 10000K
Total Submissions: 18481
 
Accepted: 8512
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

Source
IOI 1999
f송이의 꽃, v개의 꽃병을 드립니다. 꽃을 모두 꽃병에 꽂아야 합니다. 그리고 꽃의 순서는 바꿀 수 없습니다. 번호가 작은 것은 왼쪽에 있습니다. 모든 꽃을 꽃병에 넣으면 어느 정도의 가치가 있습니다. 어떻게 놓아야 가장 큰 가치가 생길 수 있는지 물어보세요.
우리는 dp[i][j]를 설정하여 i번째 꽃을 처리한 다음에 i번째 꽃을 j번째 꽃병에 넣을 때 얻을 수 있는 최대 가치를 나타낸다.
분명히 dp[i][j]=max(dp[i-1][k])+val[i][j], 그 중에서 k처음에 초기화를 잘못해서 WA를 두 번 했는데 초기화를 완전히 한 후에 AC가 됐어요.

#include <map>  
#include <set>  
#include <list>  
#include <stack>  
#include <queue>  
#include <vector>  
#include <cmath>  
#include <cstdio>  
#include <cstring>  
#include <iostream>  
#include <algorithm>  
  
using namespace std;

const int inf = -0x3f3f3f3f;
int dp[105][105];
int mat[105][105];

int main()
{
	int f, v;
	while (~scanf("%d%d", &f, &v))
	{
		memset (dp, inf, sizeof(dp));
		for (int i = 1; i <= f; ++i)
		{
			for (int j = 1; j <= v; ++j)
			{
				scanf("%d", &mat[i][j]);
			}
		}
		for (int i = 0; i <= v; ++i)
		{
			dp[0][i] = 0;
		}
		dp[1][1] = mat[1][1];
		for (int i = 1; i <= f; ++i)
		{
			for (int j = 1; j <= v; ++j)
			{
				for (int k = 1; k < j; ++k)
				{
					dp[i][j] = max(dp[i][j], dp[i - 1][k] + mat[i][j]);
				}
			}
		}
		int ans = inf;
		for (int i = 1; i <= v; ++i)
		{
			ans = max(ans, dp[f][i]);
		}
		printf("%d
", ans);
	}
	return 0;
}