思维题 UVA 10881 Piotr's Ants

题目传送门

 /*
     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
     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
                 关键2：蚂蚁的相对位置不变    关键3：order数组记录顺序
 */
 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <cstring>
 #include <string>
 #include <cmath>
 using namespace std;

 const int MAXN = 1e4 + ;
 const int INF = 0x3f3f3f3f;
 const char dir_name[][] = {"L", "Turning", "R"};
 struct Ant
 {
     int pos, dir, id;
     bool operator < (const Ant &a)    const
     {
         return pos < a.pos;
     }
 }pre[MAXN], now[MAXN];
 int order[MAXN];

 int main(void)        //UVA 10881 Piotr's Ants
 {
 //    freopen ("UVA_10881.in", "r", stdin);

     int T;    int cas = ;    int len, t, n;
     scanf ("%d", &T);
     while (T--)
     {
         scanf ("%d%d%d", &len, &t, &n);
         char ch;
         for (int i=; i<=n; ++i)
         {
             int p, d;    char ch;
             scanf ("%d %c", &p, &ch);
             d = ((ch=='L') ? - : );
             pre[i] = (Ant) {p, d, i};
             now[i] = (Ant) {p+t*d, d, };
         }

         sort (pre+, pre++n);        //计算相对位置
         for (int i=; i<=n; ++i)    order[pre[i].id] = i;    //输入(输出)的顺序

         sort (now+, now++n);
         for (int i=; i<n; ++i)
         {
             if (now[i].pos == now[i+].pos)
                 now[i].dir = now[i+].dir = ;
         }

         printf ("Case #%d:\n", ++cas);
         for (int i=; i<=n; ++i)
         {
             int x = order[i];
             if (now[x].pos <  || now[x].pos > len)    puts ("Fell off");
             else
             {
                 printf ("%d %s\n", now[x].pos, dir_name[now[x].dir+]);
             }
         }

         puts ("");
     }

     return ;
 }

 /*
 Case #1:
 2 Turning
 6 R
 2 Turning
 Fell off

 Case #2:
 3 L
 6 R
 10 R
 */