<h1 id="heading">题目</h1>
<blockquote>
<blockquote>
Piotr likes playing with ants. 
He has n of them on a horizontal pole L cm long. Each ant is facing either left or right and walks at a constant speed of 1 cm/s. 
When two ants bump into each other, they both turn around (instantaneously) and start walking in opposite directions. 
Piotr knows where each of the ants starts and which direction it is facing and wants to calculate where the ants will end up T seconds from now.
 
</blockquote>
<h2 id="input">Input</h2>
<blockquote>
The first line of input gives the number of cases, N. N test cases follow. Each one starts with a line 
containing 3 integers: L , T and n (0 ≤ n ≤ 10000). The next n lines give the locations of the n ants 
(measured in cm from the left end of the pole) and the direction they are facing (L or R).
</blockquote>
<h2 id="output">Output</h2>
<blockquote>
For each test case, output one line containing ‘Case #x:’ followed by n lines describing the locations 
and directions of the n ants in the same format and order as in the input. If two or more ants are at 
the same location, print ‘Turning’ instead of ‘L’ or ‘R’ for their direction. If an ant falls off the pole 
before T seconds, print ‘Fell off’ for that ant. Print an empty line after each test case.
</blockquote>
<h2 id="sample-input">Sample Input</h2>
<blockquote>
2 
10 1 4 
1 R 
5 R 
3 L 
10 R 
10 2 3 
4 R 
5 L 
8 R
</blockquote>
<h2 id="sample-output">Sample Output</h2>
<blockquote>
Case #1: 
2 Turning 
6 R 
2 Turning 
Fell off
Case #2: 
3 L 
6 R 
10 R
</blockquote>
</blockquote>
<h1 id="heading-1">题解</h1>
画图模拟下,很容易可以发现在忽略次序的情况下,完全可以无视碰撞,直接穿过 
单独记录次序输出即可
按照样例画下图应该不难想
给几组测试数据 
{% raw %} 
<textarea rows="5" cols="20"> 
12
10 1 2 
3 R 
4 L
10 1 2 
1 R 
2 L
10 1 2 
1 R 
3 L
3 2 2 
1 R 
3 L
3 1 2 
1 R 
3 L
10 2 3 
1 R 
2 R 
3 R
10 1 4 
1 R 
5 R 
3 L 
10 R
10 2 3 
4 R 
5 L 
8 R
1 1 1 
1 L
5 3 5 
4 L 
2 R 
5 L 
1 R 
3 L
10000 500 1 
5 R
20 5 10 
8 L 
9 L 
5 R 
1 R 
19 L 
13 R 
7 R 
16 R 
11 L 
14 R 
</textarea> 
<textarea rows="5" cols="20"> 
Case #1: 
3 L 
4 R
Case #2: 
1 L 
2 R
Case #3: 
2 Turning 
2 Turning
Case #4: 
1 L 
3 R
Case #5: 
2 Turning 
2 Turning
Case #6: 
3 R 
4 R 
5 R
Case #7: 
2 Turning 
6 R 
2 Turning 
Fell off
Case #8: 
3 L 
6 R 
10 R
Case #9: 
0 L
Case #10: 
4 R 
1 L 
5 R 
0 L 
2 L
Case #11: 
505 R
Case #12: 
6 Turning 
10 R 
4 L 
3 L 
Fell off 
14 L 
6 Turning 
19 R 
12 R 
18 R
</textarea>
{% endraw %}
<h1 id="heading-2">代码</h1>
<pre style="background-color:#fff">/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/
#include &lt;cstdio&gt;
#include &lt;algorithm&gt;
#include &lt;cmath&gt;
#include &lt;cstring&gt;
#include &lt;iomanip&gt;
#include &lt;iostream&gt;
#include &lt;map&gt;
#include &lt;set&gt;
#include &lt;list&gt;
#include &lt;queue&gt;
#include &lt;stack&gt;
#include &lt;string&gt;
#include &lt;vector&gt;
#include &lt;bitset&gt;
#include &lt;functional&gt;

using namespace std;

const int INF = 0x7FFFFFFF;
const double eps = 1e-10;

const int maxn = 10005;

struct Node {
 int n;
 int pos;
 char d;
 bool operator &lt; (const Node &amp; rhs)const {
 return pos &lt; rhs.pos;
 }
 static bool compare(Node &amp;a,Node &amp;b) {
 return a.n &lt; b.n;
 }
};
Node ants[maxn];

int kase = 1;
int num[maxn];

void Do() {
 int L,T,n;
 cin &gt;&gt; L &gt;&gt; T &gt;&gt; n;

 for(int i = 0;i &lt; n;i++) {
 cin &gt;&gt; ants[i].pos &gt;&gt; ants[i].d;
 ants[i].n = i;
 }

 sort(ants,ants + n);

 for(int i = 0;i &lt; n;i++) {
 num[ants[i].n] = i;
 if(ants[i].d == &#39;L&#39;)
 ants[i].pos -= T;
 else
 ants[i].pos += T;
 }

 sort(ants,ants + n);

 for(int i = 1;i &lt; n;i++)
 if(ants[i].pos == ants[i - 1].pos)
 ants[i].d = ants[i - 1].d = &#39;T&#39;;

 cout &lt;&lt; &#34;Case #&#34; &lt;&lt; kase++ &lt;&lt; &#34;:&#34; &lt;&lt; endl;
 for(int i = 0;i &lt; n;i++) {
 int t = num[i];
 if(ants[t].pos &gt; L || ants[t].pos &lt; 0)
 cout &lt;&lt; &#34;Fell off&#34; &lt;&lt; endl;
 else
 if(ants[t].d == &#39;T&#39;)
 cout &lt;&lt; ants[t].pos &lt;&lt; &#34; Turning&#34; &lt;&lt; endl;
 else
 cout &lt;&lt; ants[t].pos &lt;&lt; &#34; &#34; &lt;&lt; ants[t].d &lt;&lt; endl;
 }
 cout &lt;&lt; endl;

}

int main() {
 cin.tie(0);
 cin.sync_with_stdio(false);

 int T;
 cin &gt;&gt; T;
 while(T--)
 Do();

 return 0;
}
</pre>

Uva 10881.Piotr's Ants