题目传送门
1 /*
2 题意：在坐标轴上一群蚂蚁向左或向右爬，问经过ts后，蚂蚁的位置和状态
3 思维题：本题的关键1：蚂蚁相撞看作是对穿过去，那么只要判断谁是谁就可以了
4 关键2：蚂蚁的相对位置不变 关键3：order数组记录顺序
5 */
6 #include <cstdio>
7 #include <algorithm>
8 #include <iostream>
9 #include <cstring>
10 #include <string>
11 #include <cmath>
12 using namespace std;
13
14 const int MAXN = 1e4 + 10;
15 const int INF = 0x3f3f3f3f;
16 const char dir_name[][10] = {"L", "Turning", "R"};
17 struct Ant
18 {
19 int pos, dir, id;
20 bool operator < (const Ant &a) const
21 {
22 return pos < a.pos;
23 }
24 }pre[MAXN], now[MAXN];
25 int order[MAXN];
26
27 int main(void) //UVA 10881 Piotr's Ants
28 {
29 // freopen ("UVA_10881.in", "r", stdin);
30
31 int T; int cas = 0; int len, t, n;
32 scanf ("%d", &T);
33 while (T--)
34 {
35 scanf ("%d%d%d", &len, &t, &n);
36 char ch;
37 for (int i=1; i<=n; ++i)
38 {
39 int p, d; char ch;
40 scanf ("%d %c", &p, &ch);
41 d = ((ch=='L') ? -1 : 1);
42 pre[i] = (Ant) {p, d, i};
43 now[i] = (Ant) {p+t*d, d, 0};
44 }
45
46 sort (pre+1, pre+1+n); //计算相对位置
47 for (int i=1; i<=n; ++i) order[pre[i].id] = i; //输入(输出)的顺序
48
49 sort (now+1, now+1+n);
50 for (int i=1; i<n; ++i)
51 {
52 if (now[i].pos == now[i+1].pos)
53 now[i].dir = now[i+1].dir = 0;
54 }
55
56 printf ("Case #%d:
", ++cas);
57 for (int i=1; i<=n; ++i)
58 {
59 int x = order[i];
60 if (now[x].pos < 0 || now[x].pos > len) puts ("Fell off");
61 else
62 {
63 printf ("%d %s
", now[x].pos, dir_name[now[x].dir+1]);
64 }
65 }
66
67 puts ("");
68 }
69
70 return 0;
71 }
72
73 /*
74 Case #1:
75 2 Turning
76 6 R
77 2 Turning
78 Fell off
79
80 Case #2:
81 3 L
82 6 R
83 10 R
84 */