LITTLE SHOP OF FLOWERS(DP)

LITTLE SHOP OF FLOWERS
Time Limit: 1000MS
 
Memory Limit: 10000K
Total Submissions: 18300
 
Accepted: 8435
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

 
제목:
F(1~100)와 V(F~100)를 주면 F개의 화종, V개의 꽃병이 있다는 뜻이다. 현재 꽃마다 꽃병을 하나씩 골라 놓아야 한다. 꽃이 있는 꽃병에 대응하는 건강치가 있다.총화치를 최대로 하고 화종의 배치 순서를 반드시 점차적으로 늘려야 하는 방법을 묻는다.
 
아이디어:
     DP. dp[i][j]는 제i종 꽃을 j꽃병에 놓을 때 최대치이기 때문에 dp[i][j]=max {dp[i-1][k](1<=k 
     AC:

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int INF = 99999999;

int num[105][105], dp[105][105];

int main() {

    int n, m;
    scanf("%d%d", &n, &m);

    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            scanf("%d", &num[i][j]);
            if (i == 1) {
                dp[i][j] = num[i][j];
                ans = max(ans, dp[i][j]);
            }
        }
    }

    int ans = -INF;
    for (int i = 2; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            int Max = -INF;
            for (int k = 1; k < j; ++k) {
                Max = max(Max, dp[i - 1][k]);
            }

            dp[i][j] = Max + num[i][j];
            if (i == n) ans = max(ans, dp[i][j]);
        }
    }

    printf("%d
", ans);

    return 0;
}