cf B. Obsession with Robots 判断最短路？

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B. Obsession with Robots

time limit per test

2 seconds

memory limit per test

64 megabytes

input

standard input

output

standard output

The whole world got obsessed with robots,and to keep pace with the progress, great Berland's programmer Draude decided to build his own robot. He was working hard at the robot. He taught it to walk the shortest path from one point to another, to record all its movements, but like in many Draude's programs, there was a bug — the robot didn't always walk the shortest path. Fortunately, the robot recorded its own movements correctly. Now Draude wants to find out when his robot functions wrong. Heh, if Draude only remembered the map of the field, where he tested the robot, he would easily say if the robot walked in the right direction or not. But the field map was lost never to be found, that's why he asks you to find out if there exist at least one map, where the path recorded by the robot is the shortest.

The map is an infinite checkered field, where each square is either empty, or contains an obstruction. It is also known that the robot never tries to run into the obstruction. By the recorded robot's movements find out if there exist at least one such map, that it is possible to choose for the robot a starting square (the starting square should be empty) such that when the robot moves from this square its movements coincide with the recorded ones (the robot doesn't run into anything, moving along empty squares only), and the path from the starting square to the end one is the shortest.

In one movement the robot can move into the square (providing there are no obstrutions in this square) that has common sides with the square the robot is currently in.

Input

The first line of the input file contains the recording of the robot's movements. This recording is a non-empty string, consisting of uppercase Latin letters L, R, U and D, standing for movements left, right, up and down respectively. The length of the string does not exceed 100.

Output

In the first line output the only word OK (if the above described map exists), or BUG (if such a map does not exist).

Sample test(s)

input

LLUUUR

output

OK

input

RRUULLDD

output

BUG

在不知道地图什么样的情况下，给你一条走过的路径，让你判断从起点到终点的路径是否为最短距离，地图有空格也有障碍，如果有一种地图满足就输出ok。

地图不确定，路径确定，要想得到最短路，而且只要满足一种地图，而且这种地图可以自己构造，假设起点在200,200的位置，那么如果在走的过程中，同一个地点走过两遍肯定不是最短的，还有一种情况如下图，如果一个点周围不只是一个点走过，那么也不是最短路。

//30 ms	 4000 KB
#include<stdio.h>
#include<string.h>
#define M 1007
int g[M][M];
int dir[4][2]= {{0,1},{0,-1},{1,0},{-1,0}};
char s[107];
int main()
{
    while(scanf("%s",s)!=EOF)
    {
        memset(g,0,sizeof(g));
        g[200][200]=1;
        int len=strlen(s),flag=0,x=200,y=200;
        for(int i=0; i<len; i++)
        {
            if(s[i]=='L')
            {
                y--;
                if(g[x][y])
                {
                    flag=1;
                    break;
                }
                g[x][y]=1;
            }
            else if(s[i]=='R')
            {
                y++;
                if(g[x][y])
                {
                    flag=1;
                    break;
                }
                g[x][y]=1;
            }
            else if(s[i]=='U')
            {
                x--;
                if(g[x][y])
                {
                    flag=1;
                    break;
                }
                g[x][y]=1;
            }
            else if(s[i]=='D')
            {
                x++;
                if(g[x][y])
                {
                    flag=1;
                    break;
                }
                g[x][y]=1;
            }
            int cnt=0;
            for(int i=0; i<4; i++)//判断这个点周围的四个方向有几个已经走过了
            {
                int xx=x+dir[i][0];
                int yy=y+dir[i][1];
                if(g[xx][yy])cnt++;
            }
            if(cnt!=1)
            {
                flag=1;
                break;
            }
        }
        if(!flag)printf("OK\n");
        else printf("BUG\n");
    }
    return 0;
}