思维题 UVA 10881 Piotr's Ants

 

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:\n", ++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\n", 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 */