Question

C ++中用于执行以下操作的适当方法是什么：

对于此示例，允许x等于更大的int。

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

这个2D std::vector<int>就是我现在拥有的。我需要开始使用从0位置的索引开始增加的值来填充空白，以便发生以下操作：

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

继续直到填空。我查看了std::iota，但我正在使用C ++ 98（强制限制）。阅读std::generate，这似乎是一个很好的方法，不确定实施。

Answer 1

假设您将row-major order中的矩阵写入名为std::vector<int>的{{1}}中，m是一行中的位数，这将是广度的实现 - 第一搜索算法：

width

Answer 2

我的尝试并将@flyx代码转换为可以支持std::vector<std::vector<int>的表单。我正在努力进行对角线检查。

我知道缩进已关闭，直接从IDE复制。

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;
            }

我相信我已经让它大部分都在工作了，但是在输出上没有间距也难以遵循！从来没有，这是输出：

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

最后一个网格的左下角有一个部分仍然有-1个值并且没有填充。

从特定索引顺序填充2D vector <int>的方法？

2 个答案: