9

基本的に、次のような問題があります。

2D の正方配列で表現されたイチゴの庭があります。各植物（各要素）には多数のイチゴがあります。配列の左上隅から開始し、右または下にしか移動できません。庭を通る経路を計算し、どれが最も多くのイチゴを産むかを出力する再帰的な方法を設計する必要があります。

私は本当に単純な再帰問題を理解していると思いますが、この問題は私の頭をはるかに超えています。再帰メソッドの作成に関しては、どこから始めればよいのか、どこに行けばよいのかよくわかりません。

コードに関連するヘルプや、この問題の背後にある概念を理解するのに役立つヘルプは大歓迎です。ありがとう。

4

3 に答える 3

15

Like dasblinkenlight said, the most efficient way to do this is using a memoization or dynamic programming technique. I tend to prefer dynamic programming, but I'll use pure recursion here.

The answer centers around the answer to one fundamental question: "If I'm in the square in row r and column c on my field, how can I evaluate the path from the top left to here such that the number of strawberries is maximized?"

The key to realize is that there's only two ways to get in the plot in row r and column c: either I can get there from above, using the plot in row r-1 and column c, or I can get there from the side, using the plot in row r and column c-1. After that, you just need to make sure you know your base cases...which means, fundamentally, my purely recursive version would be something like:

int[][] field;    
int max(int r, int c) {
    //Base case
    if (r == 0 && c == 0) {
        return field[r][c];
    }
    //Assuming a positive number of strawberries in each plot, otherwise this needs
    //to be negative infinity
    int maxTop = -1, maxLeft = -1;
    //We can't come from the top if we're in the top row
    if (r != 0) {
        maxTop = field[r-1][c];
    }
    //Similarly, we can't come from the left if we're in the left column
    if (c != 0) {
        maxLeft = field[r][c-1];
    }
    //Take whichever gives you more and return..
    return Math.max(maxTop, maxLeft) + field[r][c];
}

Call max(r-1, c-1) to get your answer. Notice there's a lot of inefficiency here; you'll do much better by using dynamic programming (which I'll provide below) or memoization (which has already been defined). The thing to remember, though, is that both the DP and memoization techniques are simply more efficient ways that come from the recursive principles used here.

DP:

int maxValue(int[][] field) {
    int r = field.length;
    int c = field[0].length;
    int[][] maxValues = new int[r][c];
    for (int i = 0; i < r; i++) {
        for (int j = 0; j < c; j++) {
            if (i == 0 && j == 0) {
                maxValues[i][j] = field[i][j];
            } else if (i == 0) {
                maxValues[i][j] = maxValues[i][j-1] + field[i][j];
            } else if (j == 0) {
                maxValues[i][j] = maxValues[i-1][j] + field[i][j];
            } else {
                maxValues[i][j] = Math.max(maxValues[i][j-1], maxValues[i-1][j]) + field[i][j];
            }
        }
    }
    return maxValues[r-1][c-1];
}

In both cases, if you want to recreate the actual path, just keep a 2D table of booleans that corresponds with "Did I come from above or to the left"? If the most strawberry path comes from above, put true, otherwise put false. That can allow you to retrace the patch after the calculation.

Notice that this is still recursive in principal: at each step, we're looking back at our previous results. We just happen to be caching our previous results so we don't waste a bunch of work, and we're attacking the subproblems in an intelligent order so that we can always solve them. For more on dynamic programming, see Wikipedia.

于 2012-07-23T23:00:08.913 に答える
5

memoizationを使用してそれを行うことができます。以下は Java のような疑似ドキュメントです ( memo、R、およびCは、メソッドで使用可能なインスタンス変数であると想定されますmax)。

int R = 10, C = 20;
int memo[][] = new int[R][C];
for (int r=0 ; r != R ; r++)
    for (int c = 0 ; c != C ; c++)
        memo[r][c] = -1;
int res = max(0, 0, field);

int max(int r, int c, int[][] field) {
    if (memo[r][c] != -1) return memo[r][c];
    int down = 0; right = 0;
    if (r != R) down = max(r+1, c, field);
    if (c != C) right = max(r, c+1, field);
    return memo[r][c] = (field[r][c] + Math.max(down, right));
}
于 2012-07-23T22:21:15.297 に答える