<poj-2632>Crashing Robots

深藏阁楼爱情的钟 2022-07-15 07:55 211阅读 0赞

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      Crashing Robots 
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        <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
        <td>&nbsp;</td> 
        <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
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        <td><strong>Total Submissions:</strong>&nbsp;10327</td> 
        <td>&nbsp;</td> 
        <td><strong>Accepted:</strong>&nbsp;4373</td> 
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    </div> <p style="font-size:18pt; font-weight:bold; color:blue">Description</p> 
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      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.&nbsp; 
     <br> 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. 
    </div> <p style="font-size:18pt; font-weight:bold; color:blue">Input</p> 
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      The first line of input is K, the number of test cases. Each test case starts with one line consisting of two integers, 1 &lt;= A, B &lt;= 100, giving the size of the warehouse in meters. A is the length in the EW-direction, and B in the NS-direction.&nbsp; 
     <br> The second line contains two integers, 1 &lt;= N, M &lt;= 100, denoting the numbers of robots and instructions respectively.&nbsp; 
     <br> Then follow N lines with two integers, 1 &lt;= Xi &lt;= A, 1 &lt;= Yi &lt;= 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.&nbsp; 
     <br>  
     <img src="http://poj.org/images/2632_1.jpg" alt="">&nbsp; 
     <br> Figure 1: The starting positions of the robots in the sample warehouse  
     <br> Finally there are M lines, giving the instructions in sequential order.&nbsp; 
     <br> An instruction has the following format:&nbsp; 
     <br> &lt; robot #&gt; &lt; action&gt; &lt; repeat&gt;&nbsp; 
     <br> Where&nbsp;is one of&nbsp; 
     <br> 
     <ul> 
      <li>L: turn left 90 degrees,&nbsp;<br> </li> 
      <li>R: turn right 90 degrees, or&nbsp;<br> </li> 
      <li>F: move forward one meter,</li> 
     </ul> 
     <br> and 1 &lt;= &lt; repeat&gt; &lt;= 100 is the number of times the robot should perform this single move. 
    </div> <p style="font-size:18pt; font-weight:bold; color:blue">Output</p> 
    <div style="font-family:&quot;Times New Roman&quot;,Times,serif; font-size:12pt">
      Output one line for each test case:&nbsp; 
     <br> 
     <ul> 
      <li>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.)&nbsp;<br> </li> 
      <li>Robot i crashes into robot j, if robots i and j crash, and i is the moving robot.&nbsp;<br> </li> 
      <li>OK, if no crashing occurs.</li> 
     </ul> 
     <br> Only the first crash is to be reported. 
    </div> <p style="font-size:18pt; font-weight:bold; color:blue">Sample Input</p> <pre style="font-family:&quot;Courier New&quot;,Courier,monospace; font-size:12pt">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</pre> <p style="font-size:18pt; font-weight:bold; color:blue">Sample Output</p> <pre style="font-family:&quot;Courier New&quot;,Courier,monospace; font-size:12pt">Robot 1 crashes into the wall
Robot 1 crashes into robot 2
OK
Robot 1 crashes into robot 2</pre> <p style="font-size:18pt; font-weight:bold; color:blue">解题思路</p> 
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     <br> 
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    <div style="font-family:&quot;Times New Roman&quot;,Times,serif; font-size:12pt">
      就还是比较渣，程序写的十分繁琐，刚开始要做的时候，准备先去看一下解题报告还是算了 
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     <br> 
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      代码一：／／我自己的 
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     <pre><code>#include&lt;stdio.h&gt;
#include&lt;string.h&gt;

struct re
{
	int x, y;//定义ｘ，ｙ坐标
	int dri;//定义方向

}rep[100];

int main()
{
	int t, a, b, n, m;
	int nu, repeat;
	char dr, ac;
	scanf("%d",&amp;t);
	while(t--)
	{
		int flag = 1;
		memset(rep, 0, sizeof(rep));
		scanf("%d %d", &amp;a, &amp;b);
		scanf("%d %d",&amp;n, &amp;m);
		for(int i = 1; i &lt;= n; i++)
		{
			scanf("%d %d %c",&amp;rep[i].x, &amp;rep[i].y, &amp;dr);
			if(dr == 'N')
				rep[i].dri = 1;
			if(dr == 'E')
				rep[i].dri = 2;
			if(dr == 'S')
				rep[i].dri = 3;
			if(dr == 'W')
				rep[i].dri = 4;
		}
		for(int i = 1; i &lt;= m; i++)
		{
			scanf("%d %c %d", &amp;nu, &amp;ac, &amp;repeat);
			if(flag == 1)
			{
				for(int j = 1; j &lt;= repeat; j++)
				{
					if(ac == 'L')//转换方向
					{
						if(rep[nu].dri == 1)
							rep[nu].dri = 4;
						else
							rep[nu].dri--;
					}
					if(ac == 'R')
					{
						if(rep[nu].dri == 4)
							rep[nu].dri = 1;
						else
							rep[nu].dri++;
					}
					if(ac == 'F')
					{
						if(rep[nu].dri == 1)
							rep[nu].y++;
						else if(rep[nu].dri == 2)
							rep[nu].x++;
						else if(rep[nu].dri == 3)
							rep[nu].y--;
						else if(rep[nu].dri == 4)
							rep[nu].x--;

if(rep[nu].y &gt; b||rep[nu].y &lt; 1||rep[nu].x&lt;1||rep[nu].x &gt; a)
						{
							flag = 0;//标记是否发生事故
							printf("Robot %d crashes into the wall\n", nu);
							break;
						}
					}
					for(int k = 1; k &lt;= n; k++)
					{
						if(rep[nu].x == rep[k].x&amp;&amp;rep[nu].y == rep[k].y&amp;&amp;nu != k&amp;&amp;flag == 1)//除去它自身如果和他在一个格子内就crash
						{
							flag = 0;
							printf("Robot %d crashes into robot %d\n",nu, k);
							break;
						}
					}
					if(flag == 0)
						break;
				}
			}

}
		if(flag == 1)
			printf("OK\n");
	}
	return 0;
}
</code></pre> 
     <br> 代码2：　博客名字： 
     <span style="color:rgb(51,51,51); font-family:&quot;Microsoft YaHei&quot;; font-size:24px">rising_jean</span> 
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     <span style="color:rgb(51,51,51); font-family:&quot;Microsoft YaHei&quot;; font-size:24px"><br> </span> 
    </div> 
    <div style="font-family:&quot;Times New Roman&quot;,Times,serif; font-size:12pt"> 
     <span style="color:rgb(51,51,51); font-family:&quot;Microsoft YaHei&quot;; font-size:24px"></span> 
     <pre><code>#include &lt;cstdio&gt;
#include &lt;cstring&gt;
#include &lt;cstdlib&gt;
#include &lt;iostream&gt;
#include &lt;algorithm&gt;
#include &lt;cmath&gt;

using namespace std;
//0,1,2,3,N,W,S,E

int map[15][15];
char a[15][15];

int dx[] = {-1,0,1,0};
int dy[] = {0,-1,0,1};

int main()
{
    int n, m, k;
    int i, j;
    while(cin &gt;&gt; n &gt;&gt; m &gt;&gt; k){
        memset(map,0,sizeof(map));
        memset(a,0,sizeof(a));
        if(n==0&amp;&amp;m==0&amp;&amp;k==0) break;
        for(i = 0;i &lt; n;i++){
            cin &gt;&gt; a[i];
        }
        for(i = 1;i &lt;= n;i++){
            for(j = 1;j &lt;= m;j++){
                if(a[i-1][j-1] == 'N'){
                    map[i][j] = 0;
                }else if(a[i-1][j-1] == 'W'){
                    map[i][j] = 1;
                }else if(a[i-1][j-1] == 'S'){
                    map[i][j] = 2;
                }else if(a[i-1][j-1] == 'E'){
                    map[i][j] = 3;
                }
            }
        }
        int x = 1, y = k;
        int cnt[15][15];
        memset(cnt,0,sizeof(cnt));
        cnt[x][y] = 1;
        int xx =1, yy = k;
        while(1){
            x = xx;
            y = yy;
            xx = x+dx[map[x][y]];
            yy = y+dy[map[x][y]];
            if(xx&gt;=1&amp;&amp;xx&lt;=n&amp;&amp;yy&gt;=1&amp;&amp;yy&lt;=m){
                if(cnt[xx][yy] == 0){
                    cnt[xx][yy] = cnt[x][y] + 1;
                }else {
                    int tt = cnt[x][y] - cnt[xx][yy] +1;
                    printf("%d step(s) before a loop of %d step(s)\n",  cnt[xx][yy]-1, tt);
                    break;
                }
            }else {
                printf("%d step(s) to exit\n", cnt[x][y]);
                break;
            }
        }
    }

return 0;
}
</code></pre> 
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