POJ 2632 Crashing Robots

标签：poj   acm   

Description

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

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

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AC代码如下：

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

struct H
{
    int h,z;
    char f;
}rt[105];

int t;
int n,m,g,s;
int x[105],y[105];
int map[105][105];
char f[105][105];
char d[105][2];
char b;
int a,c;
int bj;

void DS()//转向
{
    c=c%4;
    if(b=='R')//左右是对称的
        c=4-c;
    if(c==1)
    {
        if(rt[a].f=='N')
            rt[a].f='W';
        else if(rt[a].f=='E')
            rt[a].f='N';
        else if(rt[a].f=='S')
            rt[a].f='E';
        else if(rt[a].f=='W')
            rt[a].f='S';
    }
    else if(c==2)
    {
        if(rt[a].f=='N')
            rt[a].f='S';
        else if(rt[a].f=='E')
            rt[a].f='W';
        else if(rt[a].f=='S')
            rt[a].f='N';
        else if(rt[a].f=='W')
            rt[a].f='E';
    }
    else if(c==3)
    {
        if(rt[a].f=='N')
            rt[a].f='E';
        else if(rt[a].f=='E')
            rt[a].f='S';
        else if(rt[a].f=='S')
            rt[a].f='W';
        else if(rt[a].f=='W')
            rt[a].f='N';
    }

}

void DF()//前进函数
{
    int i;
    if(rt[a].f=='W')
    {
        int xx=rt[a].z-c;
            for(i=rt[a].z-1;i>=xx;i--)
            if(map[rt[a].h][i]!=0&&i>=1)
              {printf("Robot %d crashes into robot %d\n",a,map[rt[a].h][i]);bj=1;break;}
              if(xx<1&&bj==0)
           {printf("Robot %d crashes into the wall\n",a);bj=1;}
            if(bj==0)
               {map[rt[a].h][rt[a].z]=0;map[rt[a].h][xx]=a;rt[a].z=xx;}
    }
    else if(rt[a].f=='N')
    {

        int xx=rt[a].h+c;
            for(i=rt[a].h+1;i<=xx;i++)
            if(map[i][rt[a].z]!=0&&i<=n)
              {printf("Robot %d crashes into robot %d\n",a,map[i][rt[a].z]);bj=1;break;}
              if(xx>n&&bj==0)
           {printf("Robot %d crashes into the wall\n",a);bj=1;}
            if(bj==0)
               {map[rt[a].h][rt[a].z]=0;map[xx][rt[a].z]=a;rt[a].h=xx;}
    }
    else if(rt[a].f=='E')
    {

        int xx=rt[a].z+c;
            for(i=rt[a].z+1;i<=xx;i++)
            if(map[rt[a].h][i]!=0&&i<=m)
              {printf("Robot %d crashes into robot %d\n",a,map[rt[a].h][i]);bj=1;break;}
              if(xx>m&&bj==0)
           {printf("Robot %d crashes into the wall\n",a);bj=1;}
            if(bj==0)
               {map[rt[a].h][rt[a].z]=0;map[rt[a].h][xx]=a;rt[a].z=xx;}

    }
    else if(rt[a].f=='S')
    {

        int xx=rt[a].h-c;
            for(i=rt[a].h-1;i>=xx;i--)
            if(map[i][rt[a].z]!=0&&i>=1)
              {printf("Robot %d crashes into robot %d\n",a,map[i][rt[a].z]);bj=1;break;}
            if(xx<1&&bj==0)
           {printf("Robot %d crashes into the wall\n",a);bj=1;}
            if(bj==0)
               {map[rt[a].h][rt[a].z]=0;map[xx][rt[a].z]=a;rt[a].h=xx;}

    }
}

int main()
{
    int i,j;
   cin>>t;
    while(t--)
    {
        bj=0;
        memset(map,0,sizeof map);
        memset(f,0,sizeof f);
        cin>>m>>n>>g>>s;
        for(i=1;i<=g;i++)
        {
            cin>>y[i]>>x[i]>>d[i][0];
            map[x[i]][y[i]]=i;
            rt[i].h=x[i];rt[i].z=y[i];
            rt[i].f=d[i][0];
            f[x[i]][y[i]]=d[i][0];
            //cout<<map[x[i]][y[i]]<<f[x[i]][y[i]]<<endl;
        }
        for(i=1;i<=s;i++)
        {
            cin>>a>>b>>c;
            if(bj==1)
                continue;
            if(b=='L')
                DS();
            else if(b=='R')
                DS();
            else if(b=='F')
                DF();
            //cout<<a<<" "<<rt[a].h<<" "<<rt[a].z<<" "<<rt[a].f<<endl<<endl;
        }
        if(bj==0)
            cout<<"OK"<<endl;
    }
    return 0;
}

POJ 2632 Crashing Robots,布布扣,bubuko.com

POJ 2632 Crashing Robots

标签：poj   acm   

原文地址：http://blog.csdn.net/hanhai768/article/details/37503063