for (int i = 0; i < agvs.size(); i++) { if (!paths[i].empty()) { Node* prev_node = nullptr; for (Node* next_node : paths[i]) { if (prev_node) { painter.drawLine(prev_node->x, prev_node->y, next_node->x, next_node->y); } prev_node = next_node; } }，一運行就卡死，修改一下

可以尝试加上一些调试语句来确定问题出在哪里，比如在循环中加上打印语句：

for (int i = 0; i < agvs.size(); i++) {
    if (!paths[i].empty()) {
        Node* prev_node = nullptr;
        for (Node* next_node : paths[i]) {
            if (prev_node) {
                painter.drawLine(prev_node->x, prev_node->y, next_node->x, next_node->y);
            }
            prev_node = next_node;
            qDebug() << "Drawing line from (" << prev_node->x << "," << prev_node->y << ") to (" << next_node->x << "," << next_node->y << ")";
        }
    }
}

这样可以看到哪一步出现了问题，进而进行调整。另外，如果是卡死的话，也可以考虑加上一些限制条件，比如最多只画几百个线段，或者加上一个定时器，每隔一段时间才画一条线段，防止一次性画太多导致卡死。

相关问题

//輸出path for (int i = 0; i < agvs.size(); i++) { for (int j = 0; j < paths[i].size(); j++) { _MAP[paths[i][j]->x][ paths[i][j]->y] = 5; } } //模擬小車行駛 for (int i = 0; i < agvs.size(); i++) { if (! paths[i].empty()) { Node* next_node = paths[i][0]; float speed = agvs[i].getSpeed(); float distance = sqrt(pow(next_node->x - agvs[i].getCurrentX(), 2) + pow(next_node->y - agvs[i].getCurrentY(), 2)); float time = distance / speed * 1000; //node_Value[next_node->x][next_node->y] = 10; QTimer::singleShot(time, this, [&, i, next_node]() { agvs[i].setCurrentX(next_node->x); agvs[i].setCurrentY(next_node->y); this->update(); // 在窗口中重绘 }); } } //如果任務完成停止定時器 for (int i = 0; i < tasks.size(); i++) { if (tasks[i].completed != 2) { allTasksCompleted = false; break; } } if (allTasksCompleted) { timer->stop(); update(); } } }，添加黨agv的current坐標輸入_MAP中，值為0

可以在第一个 for 循环中添加以下代码来实现：

for (int i = 0; i < agvs.size(); i++) {
    _MAP[agvs[i].getCurrentX()][agvs[i].getCurrentY()] = 0;
}

这样就可以将所有 AGV 的当前位置对应的地图值设为 0。完整的代码如下：

for (int i = 0; i < agvs.size(); i++) {
    for (int j = 0; j < paths[i].size(); j++) {
        _MAP[paths[i][j]->x][paths[i][j]->y] = 5;
    }
}

for (int i = 0; i < agvs.size(); i++) {
    _MAP[agvs[i].getCurrentX()][agvs[i].getCurrentY()] = 0;
}

for (int i = 0; i < agvs.size(); i++) {
    if (!paths[i].empty()) {
        Node* next_node = paths[i][0];
        float speed = agvs[i].getSpeed();
        float distance = sqrt(pow(next_node->x - agvs[i].getCurrentX(), 2) + pow(next_node->y - agvs[i].getCurrentY(), 2));
        float time = distance / speed * 1000;
        QTimer::singleShot(time, this, [&, i, next_node]() {
            agvs[i].setCurrentX(next_node->x);
            agvs[i].setCurrentY(next_node->y);
            this->update();
        });
    }
}

for (int i = 0; i < tasks.size(); i++) {
    if (tasks[i].completed != 2) {
        allTasksCompleted = false;
        break;
    }
}

if (allTasksCompleted) {
    timer->stop();
    update();
}

for (int i = 0; i < agvs.size(); i++) { if (! paths[i].empty()) { painter.setPen(QPen(Qt::red, 2)); for (int j = 0; j < paths[i].size() - 1; j++) { painter.drawLine(paths[i][j]->x, paths[i][j]->y, paths[i][j+1]->x, paths[i][j+1]->y); } } }，修改爲黨path為空時，則跳出來不運行

你可以使用 `continue` 语句来跳过当前循环迭代，直接进入下一次循环迭代。那么对于这段代码，你可以在 `if (!paths[i].empty())` 的后面添加一个 `else` 语句，并在 `else` 语句块中使用 `continue` 语句，代码如下：

for (int i = 0; i < agvs.size(); i++) {
    if (!paths[i].empty()) {
        painter.setPen(QPen(Qt::red, 2));
        for (int j = 0; j < paths[i].size() - 1; j++) {
            painter.drawLine(paths[i][j]->x, paths[i][j]->y, paths[i][j+1]->x, paths[i][j+1]->y);
        }
    } else {
        continue;
    }
}

这样，当 `paths[i]` 为空时，就会跳过内部的循环迭代。