Problem 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
***************************************************************************************************************************
大模拟，没技巧，细节很重要。
*******************************************************************************************************************************
1 #include<iostream> 2 #include<string> 3 #include<cstring> 4 #include<cstdio> 5 #include<cctype> 6 #include<queue> 7 #include<stack> 8 using namespace std; 9 struct node 10 { 11 int x,y; 12 char id; 13 }robot[1001]; 14 int a,b,n,m,case1,i,j,k; 15 int mop[1001][1001]; 16 void init() 17 { 18 cin>>a>>b>>n>>m; 19 for(i=1;i<=b;i++) 20 for(j=1;j<=a;j++) 21 mop[i][j]=0; 22 int xx,yy; 23 char de; 24 for(i=1;i<=n;i++)//此处反一下 25 { 26 cin>>xx>>yy>>de; 27 //getchar(); 28 robot[i].x=xx; 29 robot[i].y=b-yy+1; 30 robot[i].id=de; 31 //getchar(); 32 mop[b-yy+1][xx]=i; 33 } 34 35 } 36 void sovle() 37 { 38 init(); 39 int num,times; 40 char dest; 41 int cur,k=0; 42 int flag=1; 43 while(m--) 44 { 45 cin>>num>>dest>>times; 46 /*getchar(); 47 cin>>dest; 48 getchar(); 49 cin>>times; 50 */ 51 if(flag==0)continue; 52 else 53 { 54 cur=num; 55 for(i=1;i<=times;i++) 56 { 57 if(flag==0)break; 58 if(dest=='F')//直走 59 { 60 if(robot[num].id=='N') 61 { 62 mop[robot[num].y][robot[num].x]=0; 63 robot[num].y--; 64 if(robot[num].y<=0) 65 { 66 flag=0; 67 68 } 69 70 else 71 { 72 if(mop[robot[num].y][robot[num].x]==0) 73 mop[robot[num].y][robot[num].x]=num; 74 else 75 { 76 k=mop[robot[num].y][robot[num].x]; 77 flag=0; 78 } 79 } 80 } 81 else 82 if(robot[num].id=='S') 83 { 84 mop[robot[num].y][robot[num].x]=0; 85 robot[num].y++; 86 if(robot[num].y>b) 87 { 88 flag=0; 89 //break; 90 } 91 else 92 { 93 if(mop[robot[num].y][robot[num].x]!=0) 94 { 95 k=mop[robot[num].y][robot[num].x]; 96 flag=0; 97 } 98 else 99 mop[robot[num].y][robot[num].x]=num; 100 } 101 } 102 else 103 if(robot[num].id=='E') 104 { 105 mop[robot[num].y][robot[num].x]=0; 106 robot[num].x++; 107 if(robot[num].x>a) 108 { 109 flag=0; 110 //break; 111 } 112 else 113 { 114 if(mop[robot[num].y][robot[num].x]!=0) 115 { 116 k=mop[robot[num].y][robot[num].x]; 117 flag=0; 118 } 119 else 120 mop[robot[num].y][robot[num].x]=num; 121 } 122 } 123 else 124 if(robot[num].id=='W') 125 { 126 mop[robot[num].y][robot[num].x]=0; 127 robot[num].x--; 128 if(robot[num].x<=0) 129 { 130 flag=0; 131 //break; 132 } 133 else 134 { 135 if(mop[robot[num].y][robot[num].x]!=0) 136 { 137 k=mop[robot[num].y][robot[num].x]; 138 flag=0; 139 } 140 else 141 mop[robot[num].y][robot[num].x]=num; 142 } 143 } 144 } 145 else 146 if(dest=='L')//turn left 147 { 148 if(robot[num].id=='E') 149 robot[num].id='N'; 150 else 151 if(robot[num].id=='W') 152 robot[num].id='S'; 153 else 154 if(robot[num].id=='N') 155 robot[num].id='W'; 156 else 157 if(robot[num].id=='S') 158 robot[num].id='E'; 159 } 160 else 161 if(dest=='R')//turn right 162 { 163 if(robot[num].id=='E') 164 robot[num].id='S'; 165 else 166 if(robot[num].id=='W') 167 robot[num].id='N'; 168 else 169 if(robot[num].id=='N') 170 robot[num].id='E'; 171 else 172 if(robot[num].id=='S') 173 robot[num].id='W'; 174 } 175 } 176 } 177 178 } 179 if(flag==0) 180 { 181 if(k==0) 182 printf("Robot %d crashes into the wall ",cur); 183 else 184 printf("Robot %d crashes into robot %d ",cur,k); 185 } 186 else 187 printf("OK "); 188 } 189 int main() 190 { 191 cin>>case1; 192 while(case1--) 193 { 194 sovle(); 195 } 196 return 0; 197 }
坚持！！多刷题！！