POJ 2632-Crashing Robots(模拟)

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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

题意：在一个长A宽B的仓库活动，有n个机器人和m条操作，接下俩n行是机器是所在的位置以及面向的方向，接下来m行是进行的操作。判断是否撞墙、撞上其他机器人、不会碰撞。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
using namespace std;
int A,B;
int n;
struct robot {
        int x, y;
        int site;
} a[110];

int situation(int k)//判断是否发生碰撞
{
        if(a[k].x <= 0 || a[k].x > A || a[k].y <= 0 || a[k].y > B) {
                printf("Robot %d crashes into the wall\n",k);
                return 1;
        }
        for(int i = 1; i <= n; i++) {
                if(i == k)
                        continue;
                if(a[k].x == a[i].x && a[k].y == a[i].y) {
                        printf("Robot %d crashes into robot %d\n",k,i);
                        return 1;
                }
        }
        return 0;
}
int judge(char ch)//将方向数字化存储
{
        if(ch == 'E')
                return 0;
        else if(ch == 'S')
                return 1;
        else if(ch == 'W')
                return 2;
        else if(ch == 'N')
                return 3;
}

int main()
{
        int T, m,i, j;
        char ch;
        int num, cnt;
        int flag;
        scanf("%d",&T);
        while(T--) {
                scanf("%d %d %d %d",&A, &B, &n, &m);
                for(i = 1; i <= n; i++) {
                        scanf("%d %d %c",&a[i].x, &a[i].y, &ch);
                        a[i].site=judge(ch);
                }
                flag = 0;
                for(i = 0; i < m; i++) {
                        scanf("%d %c %d",&num, &ch, &cnt);
                        if(ch == 'L') {
                                a[num].site = (a[num].site - cnt % 4) % 4;
                                if(a[num].site<0)
                                        a[num].site+=4;
                        } else if(ch == 'R')
                                a[num].site = (a[num].site + cnt % 4) % 4;
                        else if(ch == 'F') {
                                if(!flag) {
                                        for(j = 0; j < cnt; j++) {
                                                if(a[num].site == 0)
                                                        a[num].x++;
                                                else if(a[num].site == 1)
                                                        a[num].y--;
                                                else if(a[num].site == 2)
                                                        a[num].x--;
                                                else if(a[num].site == 3)
                                                        a[num].y++;
                                                flag = situation (num);
                                                if(flag)
                                                        break;
                                        }
                                }
                        }
                }
                if(!flag)
                        printf("OK\n");
        }
        return 0;
}

POJ 2632-Crashing Robots(模拟)

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原文地址：http://blog.csdn.net/u013486414/article/details/42583639