C# XNA，处理大量对象，答案

【问题标题】：C# XNA, handling lots of objects,C# XNA，处理大量对象，
【发布时间】：2017-03-26 01:34:41
【问题描述】：

我正在创建一个游戏，需要一些帮助来处理一堆对象，比如大约 10000 个，在我的游戏中，我正在生成随机数量的岩石，在 1 百万乘 1 百万地图周围的随机位置，我正在将对象添加到一个列表并像那样更新和绘制它们，但运行速度很慢。我认为在这个问题上的一些帮助真的会帮助很多想要处理许多对象的学习者。这是我的生成代码：

public void WorldGeneration()
    {
        //Random Compatibility
        Random rdm = new Random();

        //Tile Variables
        int tileType;
        int tileCount = 0;
        Rock nearestRock;

        //Initialize Coordinates
        Vector2 tileSize = new Vector2(48f, 48f);
        Vector2 currentGenVector = new Vector2(48f, 48f);
        int worldTiles = 1000000;

        //Do tile generation
        for(int tile = 1; tile <= worldTiles; tile += 1)
        {   
            //Generate Classes
            tileType = rdm.Next(0, 42);
            if (tileType == 1)
            {
                if (rocks.Count != 0)
                {
                    //Check Rock Distance
                    nearestRock = rocks.FirstOrDefault(x => Vector2.Distance(x.Location, currentGenVector) < 128);
                    if (nearestRock == null)
                    {
                        Rock rock = new Rock(rockSprite, currentGenVector);
                        rocks.Add(rock);
                    }
                }
                if (rocks.Count == 0)
                {
                    Rock rock = new Rock(rockSprite, currentGenVector);
                    rocks.Add(rock);
                }
            }

            //Move Generation Tile
            if (tileCount == worldTiles / 1000)
            {
                currentGenVector.X = tileSize.X;
                currentGenVector.Y += tileSize.Y;
                tileCount = 0;
            }
            else
            {
                currentGenVector.X += tileSize.X;
            }
            //Keep Count of Tiles per layer.
            tileCount += 1;
        }

这是我的摇滚密码：

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

using Microsoft.Xna.Framework;
using Microsoft.Xna.Framework.Input;
using Microsoft.Xna.Framework.Graphics;

namespace Game2
{
    class Rock
    {
        //Draw Support
        Texture2D sprite;
        Rectangle drawRectangle;

    //Support
    Vector2 location;
    bool updating = false;

    //Active
    bool active = true;

    public Rock(Texture2D sprite, Vector2 location)
    {
        //Initialize Location/Drawing
        this.sprite = sprite;
        this.location = location;

        drawRectangle.Width = sprite.Width;
        drawRectangle.Height = sprite.Height;

        drawRectangle.X = (int)location.X - sprite.Width / 2;
        drawRectangle.Y = (int)location.Y - sprite.Height / 2;
    }

    public void Update(GameTime gameTime, MouseState mouse)
    {
        //Mining
        if (drawRectangle.Contains(mouse.X, mouse.Y))
        {
            if (mouse.LeftButton == ButtonState.Pressed)
            {
                location.X = -800;
                location.Y = -800;
            }
        }
        drawRectangle.X = (int)location.X;
        drawRectangle.Y = (int)location.Y;
    }

    public void Draw(SpriteBatch spriteBatch)
    {
        //Draws The Sprite
        spriteBatch.Draw(sprite, drawRectangle, Color.White);
    }

    //Get Location
    public Vector2 Location
    {
        get { return location; }
    }
    public bool Updating
    {
        get { return updating; }
    }
    public void setUpdating(bool updating)
    {
        this.updating = updating;
    }
    public Rectangle DrawRectangle
    {
        get { return drawRectangle; }
    }
}
}

我只是想请教一些关于如何处理所有这些对象的提示，请帮助表示赞赏

【问题讨论】：

哪个慢？更新循环还是绘制循环？您可以忽略不在当前视口中的岩石。
我认为这是更新循环，但我会试一试，我该怎么做？抱歉，我是 XNA 和 C# 的新手，还在学习。

标签： c# xna

【解决方案1】：

我处理许多对象的方法是创建一个函数来计算两个向量之间的距离，然后用它来检查当前对象是否足够接近以进行绘制。

//Distance checking code
public float GetDistance(Vector2 v1, Vector2 v2){
    float d = Math.Sqrt(Math.Abs((v1.X * v1.X) - (v2.X * v2.X)) 
    + Math.Abs((v1.Y * v1.Y) - (v2.Y * v2.Y)));
return d;
}

//example of using the Distance Check
if(GetDistance(player.position, rock.position) < 1280){
    rock.Update();
}

这只是我的想法，因此代码可能无法正常工作，但我认为这足以让您入门。祝你好运！

【讨论】：