資源分配與動態規划算法

Question

給定一組函數f1 ... fn（離散時間）和時間限制（int），應找到最大輸出，即在不同函數之間分配時間以最大化所用函數輸出的總和。

對於任何函數，任何時候的值表示如果用於所述時間的函數的總輸出。 即F（2）=函數的總輸出，如果使用2秒。 不是F（1）+ F（2）。

所有值（時間，函數輸出）都是整數。

我的當前算法通過檢查F（t）找到所有時間被放入一個函數的情況，將最大值與前一個最大M（t-1）+的所有可能輸出的最大值進行比較，找出可能損壞的最大值為每個可能的功能添加1秒（帶有已使用功能和時間的記錄）。

public int computeDamage（）{

    int totalTime = calculator.getTotalTime();
    int numAttacks = calculator.getNumAttacks();

    if(totalTime == 0) return 0;

    int[] attackHist = new int[numAttacks];

    return maxDamage(numAttacks, attackHist, 1, totalTime, 0);

}

public int maxDamage(int numAttacks, int[] attackHist, int start, int end, int max) {
    //get the max of all the values at f(start), save the attack
    int maxF = -1, attack = -1;
    for(int i = 0; i < numAttacks; i++) {
        int dam = calculator.calculateDamage(i, start);
        if(dam > maxF) {
            maxF = dam;
            attack = i;
        }
    }

    //if start isn't 1, get the max of all possible values added to the attackHist
    int maxH = -1, attackH = -1;
    if(start > 1) {
        for(int j = 0; j < numAttacks; j++) {
            int dChange = -1;
            if(attackHist[j] > 0) dChange = calculator.calculateDamage(j, attackHist[j]+1) - calculator.calculateDamage(j, attackHist[j]);
            else dChange = calculator.calculateDamage(j, attackHist[j]+1);

            if((max + dChange) > maxH) {
                maxH = max + dChange;
                attackH = j;
            }
        }

        //if max is greater, reset attackHist. Otherwise, add 1 to used attack
        if(maxF > maxH) {
            Arrays.fill(attackHist, 0);
            attackHist[attack] = start;
            max = maxF;
        } else {
            attackHist[attackH]++;
            max = maxH;
        }

    } else {
        //just set the max to maxF
        max = maxF;
        attackHist[attack] = 1;

    }

    if(end == start) return max;
    else return maxDamage(numAttacks, attackHist, start+1, end, max);
}

輸入12.in

20 12
0 3 4 7 9 12 12 14 15 15 17 19
2 5 6 9 11 12 15 15 16 19 21 22
1 4 6 8 9 11 13 14 16 18 21 22
1 4 4 4 5 5 6 8 9 11 12 14
0 3 4 5 7 10 12 13 15 17 20 20
1 3 5 5 8 10 10 12 14 15 16 18
1 1 3 5 7 8 10 11 11 13 14 16
1 1 2 2 2 3 6 7 10 11 11 12
1 3 5 5 7 7 8 11 11 12 14 16
0 1 4 5 6 9 10 11 12 12 15 18
3 5 5 7 8 10 12 12 14 15 15 16
3 5 6 9 12 12 13 14 15 18 21 21
1 2 3 4 7 9 10 12 12 15 18 18
3 4 5 7 8 10 12 13 13 16 17 20
3 5 7 7 10 11 14 16 17 18 21 23
0 1 4 7 7 8 10 12 13 13 14 16
2 3 3 6 8 9 12 15 17 18 20 21
0 2 3 3 6 8 9 10 13 15 17 17
1 2 4 7 9 9 9 11 14 14 17 19
3 5 6 7 10 11 12 12 13 16 17 19

第一行告訴我們有多少函數（20）和多少時間（12秒）來最大化輸出。

每一行都是一個定義為1到12 F（t）的函數，詳細說明了該函數在該點之前完成了多少損壞。

輸出應為31，但我的代碼輸出為30。

Answer 1

我不會嘗試調試您的代碼，但我會讓您知道它應該做什么。

你需要做的是按功能，按時間(best_value, time_this_function)創建一個表。 根據您的輸入，這是該表：

[[(0, 1), (3, 2), (4, 3), (7, 4), (9, 5), (12, 6), (12, 7), (14, 8), (15, 9), (15, 10), (17, 11), (19, 12)],
 [(2, 1), (5, 2), (6, 3), (9, 4), (11, 5), (12, 6), (15, 7), (17, 2), (18, 3), (21, 4), (23, 5), (24, 6)],
 [(2, 0), (5, 0), (6, 3), (9, 0), (11, 0), (13, 2), (15, 0), (17, 0), (19, 2), (21, 0), (23, 0), (25, 2)],
 [(2, 0), (5, 0), (6, 0), (9, 0), (11, 0), (13, 0), (15, 0), (17, 0), (19, 0), (21, 0), (23, 0), (25, 0)],
 [(2, 0), (5, 0), (6, 0), (9, 0), (11, 0), (13, 0), (15, 0), (17, 0), (19, 0), (21, 0), (23, 0), (25, 0)],
 [(2, 0), (5, 0), (6, 0), (9, 0), (11, 0), (13, 0), (15, 0), (17, 0), (19, 0), (21, 0), (23, 0), (25, 0)],
 [(2, 0), (5, 0), (6, 0), (9, 0), (11, 0), (13, 0), (15, 0), (17, 0), (19, 0), (21, 0), (23, 0), (25, 0)],
 [(2, 0), (5, 0), (6, 0), (9, 0), (11, 0), (13, 0), (15, 0), (17, 0), (19, 0), (21, 0), (23, 0), (25, 0)],
 [(2, 0), (5, 0), (6, 0), (9, 0), (11, 0), (13, 0), (15, 0), (17, 0), (19, 0), (21, 0), (23, 0), (25, 0)],
 [(2, 0), (5, 0), (6, 0), (9, 0), (11, 0), (13, 0), (15, 0), (17, 0), (19, 0), (21, 0), (23, 0), (25, 0)],
 [(3, 1), (5, 2), (8, 1), (10, 2), (12, 1), (14, 1), (16, 1), (18, 1), (20, 1), (22, 1), (24, 1), (26, 1)],
 [(3, 1), (6, 1), (8, 0), (11, 1), (13, 1), (15, 1), (17, 1), (20, 5), (22, 5), (24, 5), (26, 5), (28, 5)],
 [(3, 0), (6, 0), (8, 0), (11, 0), (13, 0), (15, 0), (17, 0), (20, 0), (22, 0), (24, 0), (26, 0), (28, 0)],
 [(3, 1), (6, 0), (9, 1), (11, 0), (14, 1), (16, 1), (18, 1), (20, 0), (23, 1), (25, 1), (27, 1), (29, 1)],
 [(3, 1), (6, 0), (9, 0), (12, 1), (14, 0), (17, 1), (19, 1), (21, 1), (23, 0), (26, 1), (28, 1), (30, 1)],
 [(3, 0), (6, 0), (9, 0), (12, 0), (14, 0), (17, 0), (19, 0), (21, 0), (23, 0), (26, 0), (28, 0), (30, 0)],
 [(3, 0), (6, 0), (9, 0), (12, 0), (14, 0), (17, 0), (19, 0), (21, 0), (23, 0), (26, 0), (28, 0), (30, 0)],
 [(3, 0), (6, 0), (9, 0), (12, 0), (14, 0), (17, 0), (19, 0), (21, 0), (23, 0), (26, 0), (28, 0), (30, 0)],
 [(3, 0), (6, 0), (9, 0), (12, 0), (14, 0), (17, 0), (19, 0), (21, 0), (23, 0), (26, 0), (28, 0), (30, 0)],
 [(3, 1), (6, 0), (9, 0), (12, 0), (15, 1), (17, 0), (20, 1), (22, 1), (24, 1), (26, 0), (29, 1), (31, 1)]]

一旦你擁有那張桌子，你需要從最后一個條目向后走，找出路徑。

每個函數花費的時間，從最后到第一個，是：

(31, 1) => 1
(28, 0) => 0
(28, 0) => 0
(28, 0) => 0
(28, 0) => 0
(28, 1) => 1
(25, 1) => 1
(22, 0) => 0
(22, 5) => 5
(10, 2) => 2
(5, 0)  => 0
(5, 0)  => 0
(5, 0)  => 0
(5, 0)  => 0
(5, 0)  => 0
(5, 0)  => 0
(5, 0)  => 0
(5, 0)  => 0
(5, 2)  => 2
# At this point we back up out of our table, and we spent no time on the first.

倒車我們得到2上的第二個功能， 2 11日， 5 12日， 1在14個， 1在最后15日和1。

希望這可以幫助您找出您缺少的步驟。