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POImprove efficiency "lighthouse"
 

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    <p>Problem Description:</p> <p>There are many light house in the sea. The range of the sea is <code>[1, 10^7] × [1, 10^7]</code>.</p> <p>Each light house can light up the south-west and north-east area. The power of the light is enought to cover any distance.</p> <p><img src="https://i.stack.imgur.com/CRYo9.png" alt="enter image description here"></p> <p>If the light house A and B are in there lighting up area, says they can illuminate each other.</p> <p>input has n+1 lines: the first line is the number of light house. the second line to the n<em>th</em> line are the horizontal and vertical coordinate of the light house.</p> <p>output has one line: the pair of light house which can illuminate each other.</p> <p>for example:</p> <p>intput:</p> <pre><code>3 2 2 4 3 5 1 </code></pre> <p>output:</p> <pre><code>1 </code></pre> <p>note:</p> <p>The coordinate of the light house is int. The horizontal and vertical coordinate of all the light houses are different.</p> <pre><code>1 ≤ x, y ≤ 10^7 </code></pre> <p>the efficiency of my code is very low. Please help me revise my code. Thank you very much!</p> <p>Here is my code. </p> <pre><code> #include&lt;stdlib.h&gt; int main() { int n,i,j,s; int *x,*y; scanf("%d",&amp;n); x=(int *)malloc(n*sizeof(int)); y=(int *)malloc(n*sizeof(int)); for(i=0;i&lt;n;i++) { scanf("%d %d",&amp;x[i],&amp;y[i]); } s=0; for(i=0;i&lt;n-1;i++) for(j=i+1;j&lt;n;j++) { if((x[i]&gt;x[j]&amp;&amp;y[i]&gt;y[j])||(x[i]&lt;x[j]&amp;&amp;y[i]&lt;y[j])) s++; } printf("%d\n",s); free(x); free(y); return 0; } </code></pre>
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    1. COAre you following Data Structures on xuetangX? Or you come across this lighthouse problem somewhere else?
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    2. COIf so, there is a special hint video on tag 第零章
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    3. COYes，its the problem on the xuetangX. I cannot solve it.
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