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  • 模拟 POJ 2632 Crashing Robots

    题目地址:http://poj.org/problem?id=2632

      1 /*
      2     题意:几个机器人按照指示,逐个朝某个(指定)方向的直走,如果走过的路上有机器人则输出谁撞到;如果走出界了,输出谁出界
      3             如果以上都没发生,则输出OK
      4     模拟题:无算法,按照条件编写几个函数和判断条件
      5     注意:1. 关于绕转,只要模拟人的转圈方向,比如向右转就向右手边转(优化:当超过4次则对4取模)
      6             2. 有先撞墙还是先撞机器人的问题(每一次'F'后都要及时判断是否OK)
      7             3. 撞哪个机器人也有先来后到的顺序('F'后以这个机器人(除自己以外)遍历一边是否OK)
      8 */
      9 #include <cstdio>
     10 #include <iostream>
     11 #include <algorithm>
     12 #include <cstring>
     13 #include <cmath>
     14 #include <string>
     15 #include <map>
     16 #include <queue>
     17 #include <vector>
     18 using namespace std;
     19 
     20 const int MAXN = 1e6 + 10;
     21 const int INF = 0x3f3f3f3f;
     22 struct NODE
     23 {
     24     int x, y;
     25     int dir;
     26 }node[MAXN];
     27 struct OP
     28 {
     29     int name, t;
     30     char move;
     31 }op[MAXN];
     32 bool flag;
     33 
     34 bool ok(int k, int N, int M, int n)
     35 {
     36     for (int i=1; i<=n; ++i)
     37     {
     38         if (node[i].x < 1 || node[i].x > N || node[i].y < 1 || node[i].y > M)
     39         {
     40             printf ("Robot %d crashes into the wall
    ", i);
     41             flag = false;
     42             return false;
     43         }
     44     }
     45 
     46     for (int j=1; j<=n; ++j)
     47     {
     48         if (k != j && node[k].x == node[j].x && node[k].y == node[j].y)
     49         {
     50             printf ("Robot %d crashes into robot %d
    ", k, j);
     51             flag = false;
     52             return false;
     53         }
     54     }
     55 
     56     flag = true;
     57     return true;
     58 }
     59 
     60 void forward(NODE *a, OP b, int N, int M, int n)
     61 {
     62     while (b.t--)
     63     {
     64         if (a->dir == 'N')    a->x -= 1;
     65         else if (a->dir == 'S')    a->x += 1;
     66         else if (a->dir == 'W')    a->y -= 1;
     67         else    a->y += 1;
     68         if (!ok (b.name, N, M, n))    break; 
     69     }
     70 }
     71 
     72 NODE turn_l(NODE a, OP b)
     73 {
     74     b.t = (b.t >= 4) ? (b.t % 4) : b.t;
     75     while (b.t--)
     76     {
     77         if (a.dir == 'N')    a.dir = 'W';
     78         else if (a.dir == 'S')    a.dir = 'E';
     79         else if (a.dir == 'E')    a.dir = 'N';
     80         else    a.dir = 'S';
     81     }
     82 
     83     return a;
     84 }
     85 
     86 NODE turn_r(NODE a, OP b)
     87 {
     88     b.t = (b.t >= 4) ? (b.t % 4) : b.t;
     89     while (b.t--)
     90     {
     91         if (a.dir == 'N')    a.dir = 'E';
     92         else if (a.dir == 'S')    a.dir = 'W';
     93         else if (a.dir == 'W')    a.dir = 'N';
     94         else    a.dir = 'S';
     95     }
     96 
     97     return a;
     98 }
     99 
    100 void work(int N, int M, int n, int m)
    101 {
    102     int i;
    103     for (i=1; i<=m; ++i)
    104     {
    105         if (op[i].move == 'F')
    106             forward (&node[op[i].name], op[i], N, M, n);
    107         else if (op[i].move == 'L')
    108             node[op[i].name] = turn_l (node[op[i].name], op[i]);
    109         else if (op[i].move == 'R')
    110             node[op[i].name] = turn_r (node[op[i].name], op[i]);
    111         if (flag)    continue;
    112         else    break;
    113     }
    114     if (i == m + 1)        puts ("OK");
    115 }
    116 
    117 int main(void)        //POJ 2632 Crashing Robots
    118 {
    119     //freopen ("G.in", "r", stdin);
    120 
    121     int t;
    122     scanf ("%d", &t);
    123     while (t--)
    124     {
    125         int N, M, n, m;
    126         scanf ("%d%d", &M, &N);
    127         scanf ("%d%d", &n, &m);
    128         for (int i=1; i<=n; ++i)
    129         {
    130             scanf ("%d %d %c", &node[i].y, &node[i].x, &node[i].dir);
    131             node[i].x = N + 1 - node[i].x;
    132         }
    133         for (int i=1; i<=m; ++i)
    134         {
    135             scanf ("%d %c %d", &op[i].name, &op[i].move, &op[i].t);
    136         }
    137 
    138         flag = true;
    139         work (N, M, n, m);
    140     }
    141 
    142     return 0;
    143 }
    144 
    145 /*
    146 Robot 1 crashes into the wall
    147 Robot 1 crashes into robot 2
    148 OK
    149 Robot 1 crashes into robot 2
    150 */
    编译人生,运行世界!
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  • 原文地址:https://www.cnblogs.com/Running-Time/p/4372497.html
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