首页 \ 问答 \ 无法理解HDU 2823的解决方案(Can't understand the solution of HDU 2823)

无法理解HDU 2823的解决方案(Can't understand the solution of HDU 2823)

 以下代码片段取自此处 。 这是HDU 2823这个问题的解决方案。  
#define eps 1e-9
double rc(point pp[],point qq[],int n,int m)    
{    
    int q=0;    
    int p=0;    
    for(int i=0;i<n;i++)     
        if(pp[i].y-pp[p].y<-eps)    
            p=i;    
    for(int i=0;i<m;i++)    
        if(qq[i].y-qq[q].y>eps)    
            q=i;    
    pp[n]=pp[0];    
    qq[m]=qq[0];    
    double tmp,ans=1e99;    
    for(int i=0;i<n;i++)    
    {    
        while((tmp=cross(pp[p+1],qq[q+1],pp[p])-cross(pp[p+1],qq[q],pp[p]))>eps)    
            q=(q+1)%m;    
        if(tmp<-eps)    
            ans=min(ans,dist_p_to_seg(qq[q],pp[p],pp[p+1]));    
        else    
            ans=min(ans,dist_seg_to_seg(pp[p],pp[p+1],qq[q],qq[q+1]));    
        p=(p+1)%n;    
    }    
    return ans;    

}    
 
 pp[]和qq[]是两个不同的凸包。 p是pp凸包的最高点， q是qq凸包的最低点。  
 我似乎无法理解这一行：  
while((tmp=cross(pp[p+1],qq[q+1],pp[p])-cross(pp[p+1],qq[q],pp[p]))>eps) 
    q=(q+1)%m;
 
 他想要实现什么目标？ 

The following code snippet was taken from here. It is the solution of this problem HDU 2823. 
#define eps 1e-9
double rc(point pp[],point qq[],int n,int m)    
{    
    int q=0;    
    int p=0;    
    for(int i=0;i<n;i++)     
        if(pp[i].y-pp[p].y<-eps)    
            p=i;    
    for(int i=0;i<m;i++)    
        if(qq[i].y-qq[q].y>eps)    
            q=i;    
    pp[n]=pp[0];    
    qq[m]=qq[0];    
    double tmp,ans=1e99;    
    for(int i=0;i<n;i++)    
    {    
        while((tmp=cross(pp[p+1],qq[q+1],pp[p])-cross(pp[p+1],qq[q],pp[p]))>eps)    
            q=(q+1)%m;    
        if(tmp<-eps)    
            ans=min(ans,dist_p_to_seg(qq[q],pp[p],pp[p+1]));    
        else    
            ans=min(ans,dist_seg_to_seg(pp[p],pp[p+1],qq[q],qq[q+1]));    
        p=(p+1)%n;    
    }    
    return ans;    

}    
 
pp[] and qq[] are two different convex hull. p is the highest point of pp convex hull and q is the lowest point of qq convex hull. 
I can't seem to understand this line:  
while((tmp=cross(pp[p+1],qq[q+1],pp[p])-cross(pp[p+1],qq[q],pp[p]))>eps) 
    q=(q+1)%m;
 
What is he trying to achieve?

原文：https://stackoverflow.com/questions/38676898

更新时间：2022-06-27 13:06

最满意答案

 是的，你可以在同一个按钮上使用onClick和onTouch，但是OnTouch回调会得到像ACTION_MOVE，ACTION_UP，ACTION_DOWN等的motionEvent，不要忘记在onTouch回调中返回false（ Details ）。 请参考下面的代码  
Button button = (Button) findViewById(R.id.button);
        button.setOnTouchListener(new OnTouchListener() {

            public boolean onTouch(View v, MotionEvent event) {
                Log.d("test", "ontouch");
                return false;
            }
        });
        button.setOnClickListener(new OnClickListener() {

            public void onClick(View v) {
                Log.d("test", "onclick");
            }

        });
 
 只需分别在TouchTouch和OnClick上执行您想要执行的操作即可。 请注意点击是用户按下按钮并释放时执行的操作，但用户按下时将触摸。  
 所以只需点击一下，日志就会像这样。 1.ACTION_DOWN，2.ACTION_UP 3. onClick  
03-22 16:19:39.735: D/test(682): ontouch
03-22 16:19:39.735: D/test(682): ontouch
03-22 16:19:39.735: D/test(682): onclick

Yes you can use both onClick and onTouch on a same button, but OnTouch callback you'll get motionEvent like ACTION_MOVE, ACTION_UP , ACTION_DOWN etc, Don't forget to return false (Details)in onTouch callback. Please refer the below code 
Button button = (Button) findViewById(R.id.button);
        button.setOnTouchListener(new OnTouchListener() {

            public boolean onTouch(View v, MotionEvent event) {
                Log.d("test", "ontouch");
                return false;
            }
        });
        button.setOnClickListener(new OnClickListener() {

            public void onClick(View v) {
                Log.d("test", "onclick");
            }

        });
 
Just do the operations you want to do in the call backs onTouch and onClick respectively. Please NOte click is a action performed when user press the button and release but Touch will be taken when user presses it. 
So on a single click the log will be like this. 1.ACTION_DOWN, 2.ACTION_UP 3. onClick 
03-22 16:19:39.735: D/test(682): ontouch
03-22 16:19:39.735: D/test(682): ontouch
03-22 16:19:39.735: D/test(682): onclick

无法理解HDU 2823的解决方案(Can't understand the solution of HDU 2823)

最满意答案

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