| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 8352 | Accepted: 3613 |
Description
In a modernized warehouse, robots are used to fetch the goods. Careful planning is needed to ensure that the robots reach their destinations without crashing into each other. Of course, all warehouses are rectangular, and all robots occupy a circular floor space with a diameter of 1 meter. Assume there are N robots, numbered from 1 through N. You will get to know the position and orientation of each robot, and all the instructions, which are carefully (and mindlessly) followed by the robots. Instructions are processed in the order they come. No two robots move simultaneously; a robot always completes its move before the next one starts moving.
A robot crashes with a wall if it attempts to move outside the area of the warehouse, and two robots crash with each other if they ever try to occupy the same spot.
Input
The first line of input is K, the number of test cases. Each test case starts with one line consisting of two integers, 1 <= A, B <= 100, giving the size of the warehouse in meters. A is the length in the EW-direction, and B in the NS-direction.
The second line contains two integers, 1 <= N, M <= 100, denoting the numbers of robots and instructions respectively.
Then follow N lines with two integers, 1 <= Xi <= A, 1 <= Yi <= B and one letter (N, S, E or W), giving the starting position and direction of each robot, in order from 1 through N. No two robots start at the same position. 
Figure 1: The starting positions of the robots in the sample warehouse
Finally there are M lines, giving the instructions in sequential order.
An instruction has the following format:
< robot #> < action> < repeat>
Where is one of
- L: turn left 90 degrees,
- R: turn right 90 degrees, or
- F: move forward one meter,
and 1 <= < repeat> <= 100 is the number of times the robot should perform this single move.
Output
Output one line for each test case:
- Robot i crashes into the wall, if robot i crashes into a wall. (A robot crashes into a wall if Xi = 0, Xi = A + 1, Yi = 0 or Yi = B + 1.)
- Robot i crashes into robot j, if robots i and j crash, and i is the moving robot.
- OK, if no crashing occurs.
Only the first crash is to be reported.
Sample Input
4 5 4 2 2 1 1 E 5 4 W 1 F 7 2 F 7 5 4 2 4 1 1 E 5 4 W 1 F 3 2 F 1 1 L 1 1 F 3 5 4 2 2 1 1 E 5 4 W 1 L 96 1 F 2 5 4 2 3 1 1 E 5 4 W 1 F 4 1 L 1 1 F 20
Sample Output
Robot 1 crashes into the wall Robot 1 crashes into robot 2 OK Robot 1 crashes into robot 2
1 #include <iostream>
2 #include<string.h>
3 using namespace std;
4
5 int main() {
6
7 int K=0;
8 int ew,ns;
9 int robots_num;
10 int instruction;
11 //读取记录机器人的位置
12 int robots_posx[105];
13 int robots_posy[105];
14 int robots_to[105];
15 //读取记录对机器人的操作
16 int ins_rob[105];
17 char ins_op[105];
18 int ins_rep[105];
19 //记录所有机器人的位置
20 int pos[100][100];
21 cin>>K;
22 //flag 记录状态 -1 Ok;0 撞到墙；其他 撞到机器人的编号
23 int flag=-1;
24 int rob_n;
25 for(int i=0;i<K;i++){
26 flag=-1;
27 rob_n=0;
28 memset(pos,0,sizeof(int)*10000);
29 cin>>ew>>ns>>robots_num>>instruction;
30 for(int j=0;j<robots_num;j++){
31 char tmp;
32 cin>>robots_posx[j]>>robots_posy[j];
33 pos[robots_posy[j]-1][robots_posx[j]-1]=j+1;
34 cin>>tmp;
35 switch(tmp){
36 case(‘N‘):
37 robots_to[j]=0;
38 break;
39 case(‘E‘):
40 robots_to[j]=1;
41 break;
42 case(‘S‘):
43 robots_to[j]=2;
44 break;
45 case(‘W‘):
46 robots_to[j]=3;
47 break;
48 }
49 }
50 for(int j=0;j<instruction;j++){
51 cin>>ins_rob[j];
52 cin>>ins_op[j];
53 cin>>ins_rep[j];
54 }
55 for(int j=0;j<instruction;j++){
56 int rob=ins_rob[j]-1;
57 char op=ins_op[j];
58 if(op==‘L‘)
59 robots_to[rob]=(robots_to[rob]+(4-ins_rep[j]%4))%4;
60 if(op==‘R‘)
61 robots_to[rob]=(robots_to[rob]+ins_rep[j])%4;
62 if(op==‘F‘){
63 int rep=ins_rep[j];
64 pos[robots_posy[rob]-1][robots_posx[rob]-1]=0;
65 int tow=robots_to[rob];
66 if(tow==0){
67 for(int k=0;k<rep;k++){
68 robots_posy[rob]++;
69 if(robots_posy[rob]>ns){
70 flag=0;
71 rob_n=rob+1;
72 break;
73 }else if(pos[robots_posy[rob]-1][robots_posx[rob]-1]!=0){
74 flag=pos[robots_posy[rob]-1][robots_posx[rob]-1];
75 rob_n=rob+1;
76 break;
77 }
78 }
79 }else if(tow==1){
80 for(int k=0;k<rep;k++){
81 robots_posx[rob]++;
82 if(robots_posx[rob]>ew){
83 flag=0;
84 rob_n=rob+1;
85 break;
86 }else if(pos[robots_posy[rob]-1][robots_posx[rob]-1]!=0){
87 flag=pos[robots_posy[rob]-1][robots_posx[rob]-1];
88 rob_n=rob+1;
89 break;
90 }
91 }
92 }else if(tow==2){
93 for(int k=0;k<rep;k++){
94 robots_posy[rob]--;
95 if(robots_posy[rob]<1){
96 flag=0;
97 rob_n=rob+1;
98 break;
99 }else if(pos[robots_posy[rob]-1][robots_posx[rob]-1]!=0){
100 flag=pos[robots_posy[rob]-1][robots_posx[rob]-1];
101 rob_n=rob+1;
102 break;
103 }
104 }
105 }else if(tow==3){
106 for(int k=0;k<rep;k++){
107 robots_posx[rob]--;
108 if(robots_posx[rob]<1){
109 flag=0;
110 rob_n=rob+1;
111 break;
112 }else if(pos[robots_posy[rob]-1][robots_posx[rob]-1]!=0){
113 flag=pos[robots_posy[rob]-1][robots_posx[rob]-1];
114 rob_n=rob+1;
115 break;
116 }
117 }
118 }
119 if(flag!=-1)
120 break;
121 pos[robots_posy[rob]-1][robots_posx[rob]-1]=rob+1;
122 }
123
124
125 }
126 if(flag==-1)
127 cout<<"OK"<<endl;
128 else if(flag==0)
129 cout<<"Robot "<<rob_n<<" crashes into the wall"<<endl;
130 else
131 cout<<"Robot "<<rob_n<<" crashes into robot "<<flag<<endl;
132 }
133 return 0;
134 }