for (int i = 0; i < grid_y; i++) { for (int j = 0; j < grid_x; j++) { float box_score = sigmoid_x(pdata[4]); ;//获取每一行的box框中含有某个物体的概率 if (box_score >= boxThreshold) { cv::Mat scores(1, className.size(), CV_32FC1, pdata + 5); Point classIdPoint; double max_class_socre; minMaxLoc(scores, 0, &max_class_socre, 0, &classIdPoint); max_class_socre = sigmoid_x(max_class_socre); if (max_class_socre >= classThreshold) { float x = (sigmoid_x(pdata[0]) * 2.f - 0.5f + j) * netStride[stride]; //x float y = (sigmoid_x(pdata[1]) * 2.f - 0.5f + i) * netStride[stride]; //y float w = powf(sigmoid_x(pdata[2]) * 2.f, 2.f) * anchor_w; //w float h = powf(sigmoid_x(pdata[3]) * 2.f, 2.f) * anchor_h; //h int left = (int)(x - 0.5 * w) * ratio_w + 0.5; int top = (int)(y - 0.5 * h) * ratio_h + 0.5; classIds.push_back(classIdPoint.x); confidences.push_back(max_class_socre * box_score); boxes.push_back(Rect(left, top, int(w * ratio_w), int(h * ratio_h))); } pdata += net_width;//下一行 } } }这段如何用neon优化，可以写段示例吗

void shiftLeft() { int i, j, k; for (i = 0; i < GRID_SIZE; i++) { for (j = 0; j < GRID_SIZE - 1; j++) { if (grid[i][j] == 0) { for (k = j + 1; k < GRID_SIZE; k++) { if (grid[i][k] != 0) { grid[i][j] = grid[i][k]; grid[i][k] = 0; break; } } } } } }

具体实现方式是，对于每一行中的数字块，从左到右遍历，若当前数字块为0，则从其右侧开始寻找第一个非0数字块，并将其移动到当前位置，同时将该数字块的原位置设置为0。例如，对于如下序列： 2 0 4 0 经过shift...

#include "path_searching/dyn_a_star.h" using namespace std; using namespace Eigen; AStar::~AStar() { for (int i = 0; i < POOL_SIZE_(0); i++) for (int j = 0; j < POOL_SIZE_(1); j++) for (int k = 0; k < POOL_SIZE_(2); k++) delete GridNodeMap_[i][j][k]; } void AStar::initGridMap(GridMap::Ptr occ_map, const Eigen::Vector3i pool_size) { POOL_SIZE_ = pool_size; CENTER_IDX_ = pool_size / 2; GridNodeMap_ = new GridNodePtr **[POOL_SIZE_(0)]; for (int i = 0; i < POOL_SIZE_(0); i++) { GridNodeMap_[i] = new GridNodePtr *[POOL_SIZE_(1)]; for (int j = 0; j < POOL_SIZE_(1); j++) { GridNodeMap_[i][j] = new GridNodePtr[POOL_SIZE_(2)]; for (int k = 0; k < POOL_SIZE_(2); k++) { GridNodeMap_[i][j][k] = new GridNode; } } } grid_map_ = occ_map; }

其中，GridMap 代表地图，GridNodeMap 表示节点列表，POOL_SIZE_ 存储节点列表的大小，CENTER_IDX_ 存储节点列表中心的索引。在 initGridMap 函数中，先设置了节点列表的大小和中心索引，然后使用 new 运算符创建...

请你解析下列代码#include <iostream>#include <vector>#include <cstdlib>#include <ctime>#include <chrono>#include <thread>class Grid {public: Grid(int width, int height) : width_(width), height_(height) { grid_.resize(width_ * height_); for (int i = 0; i < grid_.size(); ++i) { grid_[i] = rand() % 2; } } void update() { std::vector<int> new_grid(grid_.size()); for (int i = 0; i < height_; ++i) { for (int j = 0; j < width_; ++j) { int count = live_neighbors(j, i); int index = i * width_ + j; if (count == 3 || (count == 2 && grid_[index])) { new_grid[index] = 1; } else { new_grid[index] = 0; } } } grid_ = new_grid; } void print() { for (int i = 0; i < height_; ++i) { for (int j = 0; j < width_; ++j) { int index = i * width_ + j; if (grid_[index]) { std::cout << "#"; } else { std::cout << " "; } } std::cout << std::endl; } }private: int live_neighbors(int x, int y) { int count = 0; for (int j = -1; j <= 1; ++j) { for (int i = -1; i <= 1; ++i) { int col = (x + i + width_) % width_; int row = (y + j + height_) % height_; int index = row * width_ + col; count += grid_[index]; } } count -= grid_[y * width_ + x]; return count; } int width_; int height_; std::vector<int> grid_;};int main() { srand(time(nullptr)); int width, height; std::cout << "Enter grid width: "; std::cin >> width; std::cout << "Enter grid height: "; std::cin >> height; Grid grid(width, height); while (true) { grid.print(); std::this_thread::sleep_for(std::chrono::milliseconds(500)); grid.update(); } return 0;}

在 Grid 类的私有成员函数 live_neighbors(int x, int y) 中，计算细胞 (x, y) 周围存活的细胞数量。遍历以 (x, y) 为中心的 9 个细胞，并累加存活的细胞数量。最后减去细胞 (x, y) 自己的状态，得到周围存活细胞的...

int isGameOver() { int i, j; for (i = 0; i < GRID_SIZE; i++) { for (j = 0; j < GRID_SIZE; j++) { if (grid[i][j] == 0) { return 0; } if (i > 0 && grid[i][j] == grid[i - 1][j]) { return 0; } if (j > 0 && grid[i][j] == grid[i][j - 1]) { return 0; } if (i < GRID_SIZE - 1 && grid[i][j] == grid[i + 1][j]) { return 0; } if (j < GRID_SIZE - 1 && grid[i][j] == grid[i][j + 1]) { return 0; } } } return 1; }的作用

这段代码的作用是判断当前游戏是否结束。在2048游戏中，当游戏棋盘上没有空白方格且没有相邻的数字块可以合并时，游戏结束。...其中，grid是一个二维数组，存储了游戏棋盘上每个数字块的值，GRID_SIZE则是棋盘的大小。

解释代码 static int process(int8_t* input, int* anchor, int grid_h, int grid_w, int height, int width, int stride, std::vector<float>& boxes, std::vector<float>& objProbs, std::vector<int>& classId, float threshold, int32_t zp, float scale) { int validCount = 0; int grid_len = grid_h * grid_w; float thres = unsigmoid(threshold); int8_t thres_i8 = qnt_f32_to_affine(thres, zp, scale); for (int a = 0; a < 3; a++) { for (int i = 0; i < grid_h; i++) { for (int j = 0; j < grid_w; j++) { int8_t box_confidence = input[(PROP_BOX_SIZE * a + 4) * grid_len + i * grid_w + j]; if (box_confidence >= thres_i8) { int offset = (PROP_BOX_SIZE * a) * grid_len + i * grid_w + j; int8_t* in_ptr = input + offset; float box_x = sigmoid(deqnt_affine_to_f32(in_ptr, zp, scale)) 2.0 - 0.5; float box_y = sigmoid(deqnt_affine_to_f32(in_ptr[grid_len], zp, scale)) * 2.0 - 0.5; float box_w = sigmoid(deqnt_affine_to_f32(in_ptr[2 * grid_len], zp, scale)) * 2.0; float box_h = sigmoid(deqnt_affine_to_f32(in_ptr[3 * grid_len], zp, scale)) * 2.0; box_x = (box_x + j) * (float)stride; box_y = (box_y + i) * (float)stride; box_w = box_w * box_w * (float)anchor[a * 2]; box_h = box_h * box_h * (float)anchor[a * 2 + 1]; box_x -= (box_w / 2.0); box_y -= (box_h / 2.0); boxes.push_back(box_x); //push_back() 在Vector最后添加一个元素 boxes.push_back(box_y); boxes.push_back(box_w); boxes.push_back(box_h); int8_t maxClassProbs = in_ptr[5 * grid_len]; int maxClassId = 0; for (int k = 1; k < OBJ_CLASS_NUM; ++k) { int8_t prob = in_ptr[(5 + k) * grid_len]; if (prob > maxClassProbs) { maxClassId = k; maxClassProbs = prob; } } objProbs.push_back(sigmoid(deqnt_affine_to_f32(maxClassProbs, zp, scale))); classId.push_back(maxClassId); validCount++; } } } } return validCount; }

- grid_h、grid_w：特征图的高度和宽度。 - height、width：输入图像的高度和宽度。 - stride：特征图相对于输入图像的缩放因子。 - boxes：存储检测框位置和大小的向量。 - objProbs：存储目标置信度...

Cell[][] grid; int cols = 10; int rows = 10; void setup(){ size(200, 200); grid = new Cell[cols][rows]; for(int i = 0; i < cols; i ++){ for(int j = 0; j < rows; i ++){ grid[i][j] = new Cell(i20, j20, 20, 20, i+j); } } } void draw(){ background(0); for(int i = 0; i < cols; i ++){ for(int j = 0; j < rows; i ++){ grid[i][j].oscillate(); grid[i][j].display(); } } } class Cell{ float x; float y; float w; float h; float angle; Cell(float x_, float y_, float w_, float h_, float angle_){ x = x_; y = y_; w = w_; h = h_; angle = angle_; } void oscillate(){ angle += 0.02; } void display(){ stroke(255); float bright = map(sin(angle), -1, 1, 0, 255); fill(bright); rect(x, y, w, h); } }这段代码的问题怎么解决

在两个 for 循环中，第二个循环中的 i 和 j 的自增应该改为 j++，而不是 i++。所以代码应该改为： for(int i = 0; i < cols; i++){ for(int j = 0; j < rows; j++){ grid[i][j] = new Cell(i*20, j*20, 20, ...

class Solution { public int islandPerimeter(int[][] grid) { int land=0; int broad=0; for(int i=0;i<grid.length;i++){ for(int j=0;j<grid[0].length;j++){ if(grid[i][j]==1){ land++; } if(j<grid[0].length-1 || grid[i][j++]==1){ broad++; } if(i<grid.length-1 || grid[i++][j]==1){ broad++; } } } return land; } }错哪了

if (j < grid[0].length - 1 && grid[i][j + 1] == 1) { broad++; } if (i < grid.length - 1 && grid[i + 1][j] == 1) { broad++; } } } } return land * 4 - broad * 2; } } 修复后的代码会正确...

请完善该六子棋博弈树代码：#define max_depth 4 typedef struct { int x,y; }Move;//棋的坐标 int build_game_tree(GameState* state, int depth, Move *best_move) { int max_value=-100000; int min_value=100000; if (depth == 0) return 0; // 返回当前局面的评估值 for(int i=0;i<GRIDSIZE;i++) { for(int j=0;j<GRIDSIZE;j++) { if(board[i][j]==0) { int value=evaluate();//计算该位置的评估函数 if(value>max_value) { max_value=value; if(depth==max_depth) { best_move->x=i; best_move->y=j; } } } } } return best_move; }

for(int i = 0; i < GRIDSIZE; i++) { for(int j = 0; j < GRIDSIZE; j++) { if(state->board[i][j] == 0) { Move move = {i, j}; GameState* new_state = apply_move(state, move); // 生成新状态 int value ...

解释代码：static int process(int8_t* input, int* anchor, int grid_h, int grid_w, int height, int width, int stride, std::vector<float>& boxes, std::vector<float>& objProbs, std::vector<int>& classId, float threshold, int32_t zp, float scale) { int validCount = 0; int grid_len = grid_h * grid_w; float thres = unsigmoid(threshold); int8_t thres_i8 = qnt_f32_to_affine(thres, zp, scale); for (int a = 0; a < 3; a++) { for (int i = 0; i < grid_h; i++) { for (int j = 0; j < grid_w; j++) { int8_t box_confidence = input[(PROP_BOX_SIZE * a + 4) * grid_len + i * grid_w + j]; if (box_confidence >= thres_i8) { int offset = (PROP_BOX_SIZE * a) * grid_len + i * grid_w + j; int8_t* in_ptr = input + offset; float box_x = sigmoid(deqnt_affine_to_f32(in_ptr, zp, scale)) 2.0 - 0.5; float box_y = sigmoid(deqnt_affine_to_f32(in_ptr[grid_len], zp, scale)) * 2.0 - 0.5; float box_w = sigmoid(deqnt_affine_to_f32(in_ptr[2 * grid_len], zp, scale)) * 2.0; float box_h = sigmoid(deqnt_affine_to_f32(in_ptr[3 * grid_len], zp, scale)) * 2.0; box_x = (box_x + j) * (float)stride; box_y = (box_y + i) * (float)stride; box_w = box_w * box_w * (float)anchor[a * 2]; box_h = box_h * box_h * (float)anchor[a * 2 + 1]; box_x -= (box_w / 2.0); box_y -= (box_h / 2.0); int8_t maxClassProbs = in_ptr[5 * grid_len]; int maxClassId = 0; for (int k = 1; k < OBJ_CLASS_NUM; ++k) { int8_t prob = in_ptr[(5 + k) * grid_len]; if (prob > maxClassProbs) { maxClassId = k; maxClassProbs = prob; } } if (maxClassProbs>thres_i8){ objProbs.push_back(sigmoid(deqnt_affine_to_f32(maxClassProbs, zp, scale))* sigmoid(deqnt_affine_to_f32(box_confidence, zp, scale))); classId.push_back(maxClassId); validCount++; boxes.push_back(box_x); boxes.push_back(box_y); boxes.push_back(box_w); boxes.push_back(box_h); } } } } } return validCount; }

这段代码实现了一个目标检测的后处理函数，与之前的代码相比，增加了对目标置信度和类别概率的判断。具体解释如下：函数参数和功能与之前的代码相同，不再重复解释。修改部分： 1. 在检测框的置信度 box_...

请补全该六子棋博弈树代码，使其最终输出结果为下两步棋的坐标：#define max_depth 4 typedef struct { int x,y; }Move;//棋的坐标 typedef struct { int startX,startY,resultX,resultY; }res; int build_game_tree(GameState* state, int depth, Move best_move) { int max_value=-100000; int min_value=100000; if (depth == 0) return; for(int i=0;i<GRIDSIZE;i++) { for(int j=0;j<GRIDSIZE;j++) { if(board[i][j]==0) { int value=evaluate();//计算该位置的评估函数 if(value>max_value) { max_value=value; } if(depth==max_depth) { best_move->x=i; best_move->y=j; board[i][j]=1; } res.resultX=i; res.resultY=j; } } } build_game_tree(GameState state, int depth, Move *best_move) }

for(int i=0;i<GRIDSIZE;i++) { for(int j=0;j<GRIDSIZE;j++) { if(board[i][j]==0) { board[i][j]=1; // 假设此处下一步是我方棋子 int value=build_game_tree(state, depth-1, best_move); board[i][j]=0...

import java.util.ArrayList; import java.util.Collections; import java.util.Scanner; public class 1628{ public static void main(String[]args){ Scanner sc=new Scanner(System.in); int m=sc.nextInt(); int n=sc.nextInt(); int k=sc.nextInt(); int sum=0; ArrayList<ArrayList<Integer>>x=new ArrayList<>(); ArrayList<ArrayList<Integer>>y=new ArrayList<>(); for(int i=0;i<=m;i++){ x.add(new ArrayList<>()); } for(int i=0;i<=n;i++){ y.add(new ArrayList<>()); } for(int i=1;i<=m;i++){ x.get(i).add(n+1); } for(int i=1;i<=n;i++){ y.get(i).add(m+1); } for(int i=0;i<k; i++){ int a=sc.nextInt(); int b=sc.nextInt(); x.get(a).add(b); y.get(b).add(a); } for(int i=1;i<=m;i++){ int temp=0; Collections.sort(x.get(i)); for(int j=0;j<x.get(i).size();j++){ if(x.get(i).get(j)-temp>2){ sum++; } temp=x.get(i).get(j); } } for(int i=1;i<=n;i++){ int temp=0; Collections.sort(y.get(i)); for(int j=0;j<y.get(i).size();j++){ if(y.get(i).get(j)-temp>2){ sum++; }else if(y.get(i).get(j)-temp>1){ int xx=y.get(i).get(j)-1; temp=0; for(int l=0;l<x.get(xx).size();l++){ if(x.get(xx).get(l)>i){ if(x.get(xx).get(l)-temp<=2){ sum++; } break; } temp=x.get(xx).get(l); } } temp=y.get(i).get(j); } } System.out.println(sum); } }用Java换一种方式实现

for (int i = 0; i < grid.length; i++) { Arrays.fill(grid[i], 0); } // 将给定的点加入网格图中 for (int i = 0; i < k; i++) { int x = scanner.nextInt(); int y = scanner.nextInt(); grid[x][y] = 1...

逐行解释该C语言代码：#define max_depth 4 typedef struct { int x,y; }Move;//棋的坐标 typedef struct { int startX,startY,resultX,resultY; }res; int build_game_tree(GameState* state, int depth, Move *best_move) { int max_value=-100000; int min_value=100000; if (depth == 0) return evaluate(); // 返回当前局面的评估值 for(int i=0;i<GRIDSIZE;i++) { for(int j=0;j<GRIDSIZE;j++) { if(board[i][j]==0) { board[i][j]=1; // 假设此处下一步是我方棋子 int value=build_game_tree(state, depth-1, best_move); board[i][j]=0; // 恢复现场 if(value>max_value) { max_value=value; if(depth==max_depth) { best_move->x=i; best_move->y=j; } } if(value<min_value) { min_value=value; } res.resultX=i; res.resultY=j; } } } if (depth == max_depth) { printf("下一步应该落子的位置：(%d,%d)\n", best_move->x, best_move->y); } return (state->player == 1) ? max_value : min_value; // 返回当前局面的最优值 }

for(int i=0;i<GRIDSIZE;i++) { for(int j=0;j<GRIDSIZE;j++) { if(board[i][j]==0) { board[i][j]=1; int value=build_game_tree(state, depth-1, best_move); board[i][j]=0; if(value>max_value) { max...

for (int i = 0; i < score_table.size(); i++) { if (score_table[i] > detection_threshold_) { cv::Point2f pos = feature_table[i]; features.push_back(cv::Point2f(pos.x, pos.y)); } } zero_occupancy_grid();

这段代码是在遍历之前更新过的 score_table 分数表，将分数高于指定阈值 detection_threshold_ 的网格对应的特征点...最后，调用 zero_occupancy_grid() 函数将占用网格清零，以便下一次特征点检测时再次使用。

private void ResetGrid() { for (int i = 0; i < max; i++) { for (int j = 0; j < max; j++) { if (grid[i, j] != -1) { grid[i, j] = -1; canvas1.Children.Remove(images[i, j]); images[i, j] = null; } } } }

具体实现是通过两个嵌套的 for 循环遍历整个网格，如果当前格子的值不为 -1，则将其值设为 -1，并从画布中移除对应的图片。最后将图片的引用设为 null，以便下一次重新加载图片。这个方法的作用可能是在游戏中需要...

var arr = BinaryDataMgr.Instance.GetData("Height_" + levelId); int index = 0; Vector3 startPos = new Vector3(0, 0, 0); for (int i = 0; i < length; i++) { Vector3 pos = startPos; for (int j = 0; j < width; j++) { GameObject game = new GameObject(); game.transform.SetParent(GameObject.Find("Map").transform); game.transform.localPosition = pos; game.transform.localScale = new Vector3(Gridsize, arr[index] / 1000.0f, Gridsize); game.AddComponent<DrawHeightGround>(); pos.x += Gridsize; pos.y = 4; index++; } startPos.z += Gridsize; } 这段代码如何改成从左上角(0,0,0,)开始遍历格子

可以将 startPos 初始化为左上角的位置，即 (0, 0, 0)，然后在 for 循环中，先对 j 进行遍历，再对 i 进行遍历，这样就可以从左上角开始遍历格子了。修改后的代码如下： var arr = BinaryDataMgr....

void NCut::GetPatchFeature(const MatrixXf &points, MatrixXf features_in) { MatrixXf &features = features_in; const int num_points = static_cast<int>(points.rows()); const int dim = _patch_size * _patch_size; features.resize(num_points, dim); for (int i = 0; i < num_points; ++i) { // .1 patch int irow = 0; int jcol = 0; _ff_feature_grid.Pos2d(points.coeffRef(i, 0), points.coeffRef(i, 1), &irow, &jcol); cv::Mat patch; cv::Point2f pt(static_cast<float>(irow), static_cast<float>(jcol)); cv::getRectSubPix(_cv_feature_map, cv::Size(_patch_size, _patch_size), pt, patch); // .2 maybe later i will add other features based on patch // .3 add to features assert(dim == patch.rows * patch.cols); int p = 0; for (int r = 0; r < patch.rows; ++r) { for (int c = 0; c < patch.cols; ++c) { float val = patch.at<float>(r, c); features.coeffRef(i, p++) = static_cast<float>( (std::isnan(val) || std::isinf(val)) ? 1.e-50 : val); // features.coeffRef(i, p++) = patch.at<float>(r, c); } } } }

给定一个点集points，对于每个点，首先通过_ff_feature_grid将其映射到_feature_map图像上，并取该点为中心的_patch_size * _patch_size大小的矩形区域作为图像块。然后，对于每个图像块，将像素值添加到特征向量中...

for (int i = 0; i < 60; i++) { for (int j = 0; j < 60; j++) { a[i][j] = i*j; } 请对上述代码进行并行计算，我的平台是VS2022以及CUDA11.6。分为三个线程块计算，第一个线程计算i从0-19，第二个线程计算i从20-39，第三个线程计算i从40-59。

for (int j = 0; j < 60; j++) { a[x][j] = x * j; } } int main() { int a[60][60]; int (*dev_a)[60]; cudaMalloc((void **)&dev_a, sizeof(int) * 60 * 60); dim3 blockSize(20, 1, 1); dim3 gridSize...

class Solution { public int maxValue(int[][] grid) { int r = grid.length; int c = grid[0].length; int[][] f = new int[grid.length + 1][grid[0].length +1]; for(int i = 1; i <= r; i++){ for(int j = 1 ; j <= c ; j++){ f[i][j] = Math.max(f[i-1][j],f[i][j-1])+grid[i-1][j-1]; // System.out.print(f[i][j] + " "); } } return f[r][c]; } }改成C语言的形式

for (int i = 0; i < r; i++) { for (int j = 0; j < c; j++) { scanf("%d", &grid[i][j]); } } int maxVal = maxValue(grid, r, c); printf("网格中从左上角走到右下角可以获得的最大价值为：%d\n", maxVal...

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void shiftLeft() { int i, j, k; for (i = 0; i < GRID_SIZE; i++) { for (j = 0; j < GRID_SIZE - 1; j++) { if (grid[i][j] == 0) { for (k = j + 1; k < GRID_SIZE; k++) { if (grid[i][k] != 0) { grid[i][j] = grid[i][k]; grid[i][k] = 0; break; } } } } } }

for (int i = 0; i < score_table.size(); i++) { if (score_table[i] > detection_threshold_) { cv::Point2f pos = feature_table[i]; features.push_back(cv::Point2f(pos.x, pos.y)); } } zero_occupancy_grid();

private void ResetGrid() { for (int i = 0; i < max; i++) { for (int j = 0; j < max; j++) { if (grid[i, j] != -1) { grid[i, j] = -1; canvas1.Children.Remove(images[i, j]); images[i, j] = null; } } } }

for (int i = 0; i < 60; i++) { for (int j = 0; j < 60; j++) { a[i][j] = i*j; } 请对上述代码进行并行计算，我的平台是VS2022以及CUDA11.6。分为三个线程块计算，第一个线程计算i从0-19，第二个线程计算i从20-39，第三个线程计算i从40-59。

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