Crashing Robots - poj 2632

时间：2015-07-05 12:20:38 阅读：93 评论：0 收藏：0 [点我收藏+]

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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.
技术分享

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

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 }

Crashing Robots - poj 2632

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原文地址：http://www.cnblogs.com/sdxk/p/4620403.html

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