順序填充二維矢量的方法 <int> 從特定索引？

Question

什么是C ++中執行以下操作的適當方法：

對於此示例，允許x等於更大的int 。

xxxxxxxxxxxxxxxxxx
xx    xxxxxx0000xx
xx     xxxxx    xx
xx     xxxxx    xx
xx              xx
xx              xx
xxxxxxxxxxxxxxxxxx

這是我目前擁有的2d std::vector<int> 。 我需要開始從0的索引開始以遞增的值填充空白，以便進行以下操作：

xxxxxxxxxxxxxxxxxx
xx    xxxxxx0000xx
xx   99xxxxx1111xx
xx   98xxxxx2222xx
xx   98765433333xx
xx   98765444444xx
xxxxxxxxxxxxxxxxxx

並繼續直到所有空格都被填充。 我看了std::iota但是我在C ++ 98上運行（施加了限制）。 閱讀有關std::generate ，這似乎是一個不錯的方法，不確定實現。

Answer 1

假設您以行優先順序將矩陣寫入名為m的std::vector<int>並且width是一行中的位數，則這將是廣度優先搜索算法的實現：

// these are the x values
#define OBSTACLE -1
// these are empty fields
#define EMPTY -2

std::vector<int> m;
int width;

// TODO: fill the vector here with OBSTACLEs and EMPTY and set the width!

// the list contains indexes of filled, yet unprocessed fields
std::queue<int> worklist;

// now initialize the starting zeroes. I am using the indexes
// from your example matrix here.
worklist.push(1 * width + 12); m[1 * width + 12] = 0; // second row, 13th field
worklist.push(1 * width + 13); m[1 * width + 13] = 0; // second row, 14th field
worklist.push(1 * width + 14); m[1 * width + 14] = 0; // .. and so on.
worklist.push(1 * width + 15); m[1 * width + 15] = 0;

// algorithm
while (worklist.size() > 0) {
    const int cur = worklist.front();
    worklist.pop();
    const int newval = m[cur] + 1;
    int directions[4] = {0, 0, 0, 0};
    if (cur >= width) {
        // we are not on the first row, check upwards
        directions[0] = -width;
    }
    if (cur % width != 0) {
        // we are not on the first column, check left
        directions[1] = -1;
    }
    if ((cur + 1) % width != 0) {
        // we are not on the last column, check right
        directions[2] = 1;
    }
    if (cur + width < m.size()) {
        // we are not on the last row, check downwards
        directions[3] = width;
    }
    // I am not calculating diagonal connections here; you also seem to
    // want those. They would be trivial to add.
    for (int i = 0; i < 4; ++i) {
        if (directions[i] != 0) {
            const int index = cur + directions[i];
            if (m[index] == EMPTY) {
                m[index] = newval;
                worklist.push(index);
            }
        }
    }
}

Answer 2

我試圖將@flyx代碼轉換為可以支持std::vector<std::vector<int> 。 我在對角線支票上掙扎。

我知道縮進已關閉，可以直接從IDE復制。

queue<pair<int, int> > worklist;

            for (size_t i = 0; i < nlevel.size(); i++) {
                for (size_t j = 0; j < nlevel[i].size(); j++) {
                    if (nlevel[i][j] == 0) {
                        /*cout << i << ' ' << j << endl;*/
                        worklist.push(make_pair(i,j));
                    }
                }
            }

            while (worklist.size() > 0) {
                pair<int, int> cur = worklist.front();
                worklist.pop();
                const int newval = nlevel[cur.first][cur.second] + 1;
                int directions[4] = {0, 0, 0, 0};
                int width = nlevel[0].size();
                int height = nlevel.size();
                if (cur.first > 0) {
                    // we are not on the first row, check upwards
                    directions[0] = -1;
                }
                if (cur.second > 0) {
                    // we are not on the first column, check left
                    directions[2] = -1;
                }
                if (cur.first < height) {
                    // we are not on the last row, check downwards
                    directions[1] = 1;
                }
                if (cur.second < width) {
                    // we are not on the last column, check right
                    directions[3] = 1;
                }
                for (int i = 0; i < 2; i++) {
                    for (int j = 2; j < 4; j++) {
                        if (directions[i] != 0 && directions[j] != 0) {
                            const int xindex = cur.first + directions[j];
                            const int yindex = cur.second + directions[i];
                            //cout << xindex << ' ' << yindex << endl;
                            if (nlevel[xindex][yindex] == -1) {
                                nlevel[xindex][yindex] = newval;
                                worklist.push(make_pair(xindex, yindex));
                            }
                        }
                    }

                }
            }

            cout << endl;

            for (size_t i = 0; i < nlevel.size(); i++) {
                for (size_t j = 0; j < nlevel[i].size(); j++) {
                    cout << nlevel[i][j];
                }
                cout << endl;
            }

我相信我已經使它大部分工作了，但是在輸出上沒有間距卻很難遵循！ 從來沒有少，這是輸出：

xxxxXxxxxXxxxxXxxxxXxxxxXxxxxX
xxxxxx      xxxxx      xxx  xx
xxxxx        xxx       xx    x
xx     xx     x        x     x
X      xx              x  x  x
x      xx                 x  x
x      xxxxxxxxxxxxxx     x  x
x     xxxxxxxxxxxxxxxxxxxxx  x
xFFFF x  xxxxxxxxxxxxxx      x
XFFFF x    xxxxxxxx          x
x     x      1               x
xxxxxxx      2        xxx    x
xxxxxx       3         x     x
x                            x
Xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2
-2-2-2-2-2-2-1-1-1-1-1-1-2-2-2-2-2-1-1-1-1-1-1-2-2-2-1-1-2-2
-2-2-2-2-2-1-1-1-1-1-1-1-1-2-2-2-1-1-1-1-1-1-1-2-2-1-1-1-1-2
-2-2-1-1-1-1-1-2-2-1-1-1-1-1-2-1-1-1-1-1-1-1-1-2-1-1-1-1-1-2
-2-1-1-1-1-1-1-2-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-2-1-1-2
-2-1-1-1-1-1-1-2-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-2
-2-1-1-1-1-1-1-2-2-2-2-2-2-2-2-2-2-2-2-2-2-1-1-1-1-1-2-1-1-2
-2-1-1-1-1-1-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-1-1-2
-20000-1-2-1-1-2-2-2-2-2-2-2-2-2-2-2-2-2-2-1-1-1-1-1-1-2
-20000-1-2-1-1-1-1-2-2-2-2-2-2-2-2-1-1-1-1-1-1-1-1-1-1-2
-2-1-1-1-1-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2
-2-2-2-2-2-2-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-2-2-1-1-1-1-2
-2-2-2-2-2-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-1-1-1-2
-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2
-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2

-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2
-2-2-2-2-2-27779911-2-2-2-2-2171719192121-2-2-22727-2-2
-2-2-2-2-266688101012-2-2-216161818202022-2-226262628-2
-2-255555-2-299111113-21515171719192121-22525252727-2
-2444444-2-21010101212141416161818202022-22424-22628-2
-2333333-2-21111111113131515171719192121232325-22927-2
-2222222-2-2-2-2-2-2-2-2-2-2-2-2-2-22022222424-22830-2
-211111-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-23129-2
-200001-25050-2-2-2-2-2-2-2-2-2-2-2-2-2-2343432323032-2
-200001-251494947-2-2-2-2-2-2-2-239373735353333313331-2
-211111-250504848464644444242404038383636343432343234-2
-2-2-2-2-2-2-2514949474745454343414139393737-2-2-235333533-2
-2-2-2-2-2-25250504848464644444242404038383838-23634363436-2
-2-1-1-1-1535151494947474545434341413939393939373537353735-2
-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2

最后一個網格的左下方有一個部分，該部分仍然具有-1值並且沒有填充。