模拟 --- Crashing Robots
Crashing Robots
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 7394 | Accepted: 3242 |
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.
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
and 1 <= < repeat> <= 100 is the number of times the robot should perform this single move.
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:
Only the first crash is to be reported.
- 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
【题目来源】
http://poj.org/problem?id=2632
【题目大意】
poj 2632
在一个大型仓库里，机器人常常需要来往取货。需要一系列的指令来操作机器人使得每个机器人移动到他的目的地。并且不能撞到其他的机器人。当然，所有的仓库都是矩形的，每个的机器人占据一个半径为1的圆形。
假定有N个机器人，编号为1到N。你将会知道每个机器人的位置和方向，并且所有下达的指令，机器人将会听从这些指令。指令出来后将会被处理。
没有两个机器人将会同时移动，每个机器人移动完后下一个机器人才会移动。
一个机器人撞到墙壁的话，他将会被撞毁；如果两个机器人相撞的话，两个机器人都将会被撞毁。
输入：
第一行是一个整数K，表示测试情况有K个。
每个测试情况的第一行是两个整数，1<=A,B<=100，表示这个仓库的大小。A表示横向的长度，B表示纵向的长度。
第二行有两个数，1<=N,M<=100,分别表示机器人的数量和指令数目。
接下来有N行，每行有两个数字和一个字母(N,S,E,W)，表示每一个机器人的坐标和方向。没有相同的两个机器人在相同的位置上。
最后是M行，代表一连串的指令。
每个指令的格式如下：
<哪个机器人><要执行的动作><重复的次数>
其中动作包括：
L：向左转90度
R：向右转90度
F：向前走
每次重复的次数大于等于1小于等于10。
输出：
每个test case输出1行。
输出包括：
如果机器人i撞到了墙壁，那么输出：Robot i crashes into the wall
如果机器人i撞到了机器人j，那么输出：Robot i crashes into robot j
如果没有机器人相撞，那么输出：OK
只需要报告第一次撞击。
【题目分析】
比较简单的长模拟，就是代码长，估计是被我写残了，写了250多行。做的时候要注意细节，比如说这题的地图的y轴和数组是反的，做的时候要反过来，还有横纵坐标也是反的。注意了这些细节就可以ac了。
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<cmath> #include<algorithm> #define MAX 150 using namespace std; struct Node { bool have; int dir; int number; }; Node Map[MAX][MAX]; int A,B; struct Order { int num; char op; int time; }; Order order[MAX]; void make_Map(int n,int m) { int i,j; for(i=0;i<=n+1;i++) { for(j=0;j<=m+1;j++) { Map[i][j].have=false; Map[i][j].dir=0; Map[i][j].number=0; } } //将地图围起来，这里可以使用dir这个值来标记围墙 for(i=0;i<=n+1;i++) Map[i][0].dir=Map[i][m+1].dir=-1; for(i=0;i<=m+1;i++) Map[0][i].dir=Map[n+1][i].dir=-1; // for(i=0;i<=n+1;i++) // { // for(j=0;j<=m+1;j++) // { // printf("%d ",Map[i][j].dir); // } // puts(""); // } } bool go(int num,char op,int time) { // printf("num=%d op=%c time=%d\n",num,op,time); int i,j; int x1,y1; int number1,dir1; // for(i=1;i<=B;i++) // { // for(j=1;j<=A;j++) // { // printf("%d ",Map[i][j].number); // } // puts(""); // } for(i=1;i<=B;i++) { for(j=1;j<=A;j++) { if(Map[i][j].number==num) { x1=i; y1=j; break; } } } // printf("x1=%d y1=%d\n",x1,y1); if(op=='L') //右转90度 { for(i=1;i<=time;i++) { Map[x1][y1].dir=(Map[x1][y1].dir+1)%4; } } else if(op=='R') //左转90度 { for(i=0;i<time;i++) { Map[x1][y1].dir=(Map[x1][y1].dir-1); if(Map[x1][y1].dir==0) Map[x1][y1].dir=4; } } else //前进 { // printf("x1=%d y1=%d\n",x1,y1); // printf("Map[%d][%d].dir=%d\n",x1,y1,Map[x1][y1].dir); switch(Map[x1][y1].dir) { case 1: //往上走 number1=Map[x1][y1].number; dir1=Map[x1][y1].dir; Map[x1][y1].number=0; Map[x1][y1].have=!Map[x1][y1].have; // printf("This is case 1\n"); for(i=0;i<time;i++) { x1--; if(Map[x1][y1].have) //遇到了其他机器人 { printf("Robot %d crashes into robot %d\n",number1,Map[x1][y1].number); return true; } if(Map[x1][y1].dir==-1)//撞到了墙 { printf("Robot %d crashes into the wall\n",number1); return true; } } Map[x1][y1].have=true; Map[x1][y1].dir=dir1; Map[x1][y1].number=number1; break; case 2: //往右走 number1=Map[x1][y1].number; dir1=Map[x1][y1].dir; Map[x1][y1].number=0; Map[x1][y1].have=!Map[x1][y1].have; // printf("This is case 2\n"); for(i=0;i<time;i++) { y1++; if(Map[x1][y1].have) //遇到了其他机器人 { printf("Robot %d crashes into robot %d\n",number1,Map[x1][y1].number); return true; } if(Map[x1][y1].dir==-1)//撞到了墙 { printf("Robot %d crashes into the wall\n",number1); return true; } } Map[x1][y1].have=true; Map[x1][y1].dir=dir1; Map[x1][y1].number=number1; break; case 3://往下走 number1=Map[x1][y1].number; dir1=Map[x1][y1].dir; Map[x1][y1].number=0; Map[x1][y1].have=!Map[x1][y1].have; // printf("This is case 3\n"); for(i=0;i<time;i++) { x1++; if(Map[x1][y1].have) //遇到了其他机器人 { printf("Robot %d crashes into robot %d\n",number1,Map[x1][y1].number); return true; } if(Map[x1][y1].dir==-1)//撞到了墙 { printf("Robot %d crashes into the wall\n",number1); return true; } } Map[x1][y1].have=true; Map[x1][y1].dir=dir1; Map[x1][y1].number=number1; break; case 4://往左走 number1=Map[x1][y1].number; dir1=Map[x1][y1].dir; Map[x1][y1].number=0; Map[x1][y1].have=!Map[x1][y1].have; // printf("This is case 4\n"); for(i=0;i<time;i++) { y1--; if(Map[x1][y1].have) //遇到了其他机器人 { printf("Robot %d crashes into robot %d\n",number1,Map[x1][y1].number); return true; } if(Map[x1][y1].dir==-1)//撞到了墙 { printf("Robot %d crashes into the wall\n",number1); return true; } } Map[x1][y1].have=true; Map[x1][y1].dir=dir1; Map[x1][y1].number=number1; break; } } return false; } int main() { freopen("in.txt","r",stdin); freopen("out.txt","w",stdout); int T; cin>>T; while(T--) { cin>>A>>B; make_Map(B,A);//注意，题目给的方向要反过来 int N,M; cin>>N>>M; int i,j,k; int x1,y1; char c; for(i=0;i<N;i++) { scanf("%d %d %c",&y1,&x1,&c); Map[x1][y1].have=true; if(c=='S') Map[x1][y1].dir=1; else if(c=='E') Map[x1][y1].dir=2; else if(c=='N') Map[x1][y1].dir=3; else Map[x1][y1].dir=4; Map[x1][y1].number=i+1; // cout<<Map[x1][y1].number<<endl; // cout<<"x1="<<x1<<" "<<"y1="<<y1<<" "<<"c="<<c<<endl; } int number1; char op; int time; for(i=0;i<M;i++) { scanf("%d %c %d",&order[i].num,&order[i].op,&order[i].time); // printf("%d %c %d\n",order[i].num,order[i].op,order[i].time); } bool flag=false; for(i=0;i<M;i++) { // printf("num=%d op=%c time%d\n",order[i].num,order[i].op,order[i].time); if(go(order[i].num,order[i].op,order[i].time)) { flag=true; break; } } if(!flag) printf("OK\n"); } return 0; }