#include #include #include using namespace std;

// 计算连通分量的个数 int countComponents(vector<vector>& grid) { int count = 0; int rows = grid.size(); int cols = grid[0].size(); vector<vector> visited(rows, vector(cols, false)); vector<vector> dirs{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

for (int i = 0; i < rows; i++) {
    for (int j = 0; j < cols; j++) {
        if (grid[i][j] == '+' && !visited[i][j]) {
            count++;
            queue<pair<int, int>> q;
            q.push({i, j});
            visited[i][j] = true;
            
            while (!q.empty()) {
                int x = q.front().first;
                int y = q.front().second;
                q.pop();
                
                for (auto dir : dirs) {
                    int nx = x + dir[0];
                    int ny = y + dir[1];
                    if (nx >= 0 && nx < rows && ny >= 0 && ny < cols && grid[nx][ny] == '+' && !visited[nx][ny]) {
                        q.push({nx, ny});
                        visited[nx][ny] = true;
                    }
                }
            }
        }
    }
}

return count;

}

// 布线格子的数量 int countWires(vector<vector>& grid) { int count = 0; int rows = grid.size(); int cols = grid[0].size();

for (int i = 0; i < rows; i++) {
    for (int j = 0; j < cols; j++) {
        if (grid[i][j] == '+') {
            count++;
        }
    }
}

return count;

}

// DFS搜索连通分量,并标记为布线 void dfs(vector<vector>& grid, vector<vector>& visited, int x, int y) { int rows = grid.size(); int cols = grid[0].size(); vector<vector> dirs{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

visited[x][y] = true;
grid[x][y] = '+';

for (auto dir : dirs) {
    int nx = x + dir[0];
    int ny = y + dir[1];
    if (nx >= 0 && nx < rows && ny >= 0 && ny < cols && grid[nx][ny] == '.' && !visited[nx][ny]) {
        dfs(grid, visited, nx, ny);
    }
}

}

// 尽可能多的布线 void layoutRouting(vector<vector>& grid) { int rows = grid.size(); int cols = grid[0].size(); vector<vector> visited(rows, vector(cols, false));

// DFS搜索连通分量,并标记为布线
for (int i = 0; i < rows; i++) {
    for (int j = 0; j < cols; j++) {
        if (grid[i][j] == '.') {
            dfs(grid, visited, i, j);
            return;
        }
    }
}

}

int main() { int n, m; cin >> n >> m; vector<vector> grid(n, vector(m));

for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
        cin >> grid[i][j];
    }
}

layoutRouting(grid);

for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
        cout << grid[i][j];
    }
    cout << endl;
}

return 0;
题目描述电路布局布线Circuit Layout and Routing是电子设计自动化EDA领域的一个重要概念它涉及到在电路板或集成电路上安排和连接电子元件的过程。这个过程的目标是在满足电气性能、信号完整性、电磁兼容性等要求的同时实现对空间、成本和生产工艺的优化。小小现在需要解决一个简化的电路布线问题在一个 �×�n×m 的方格中进行电路布线。其中:井号 # 标记的格子已经被占用不能布线。加号

原文地址: http://www.cveoy.top/t/topic/hRkw 著作权归作者所有。请勿转载和采集!

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