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最新推荐文章于 2023-04-20 16:42:06 发布

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本文介绍了一个关于海豚在二维平面上寻找从起点到终点再返回的最短路径问题。海豚需要两次往返于两个点之间，并且除了起始点和终点外不能重复经过同一个点。文章提供了具体的输入输出示例及一种可能的解决方案。

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Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up.

Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1.
Here, both the x- and y-coordinates before and after each movement must be integers.
He will first visit the point (tx,ty) where sx<tx and sy<ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy).
Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty).
Under this condition, find a shortest path for him.Constraints

−1000≤sx<tx≤1000
−1000≤sy<ty≤1000
sx,sy,tx and ty are integers.

Input

The input is given from Standard Input in the following format:

sx sy tx ty

Output

Print a string S that represents a shortest path for Dolphin.
The i-th character in S should correspond to his i-th movement.
The directions of the movements should be indicated by the following characters:

U: Up
D: Down
L: Left
R: Right

If there exist multiple shortest paths under the condition, print any of them.

Sample Input 1

0 0 1 2

Sample Output 1

UURDDLLUUURRDRDDDLLU

One possible shortest path is:

Going from (sx,sy) to (tx,ty) for the first time: (0,0) → (0,1) → (0,2) → (1,2)
Going from (tx,ty) to (sx,sy) for the first time: (1,2) → (1,1) → (1,0) → (0,0)
Going from (sx,sy) to (tx,ty) for the second time: (0,0) → (−1,0) → (−1,1) → (−1,2) → (−1,3) → (0,3) → (1,3) → (1,2)
Going from (tx,ty) to (sx,sy) for the second time: (1,2) → (2,2) → (2,1) → (2,0) → (2,−1) → (1,−1) → (0,−1) → (0,0)

Sample Input 2

-2 -2 1 1

Sample Output 2

UURRURRDDDLLDLLULUUURRURRDDDLLDL

AC:

#include<stdio.h>

int main()
{
int a,b,c,d;
int m,n,i;
scanf("%d%d%d%d",&a,&b,&c,&d);
m=d-b;n=c-a;
for(i=0;i<m;i++)
printf("U");
for(i=0;i<n;i++)
printf("R");
for(i=0;i<m;i++)
printf("D");
for(i=0;i<n;i++)
printf("L");
int a1=a-1;
int d1=d+1;
printf("L");
m=d1-b;n=c-a1;
for(i=0;i<m;i++)
printf("U");
for(i=0;i<n;i++)
printf("R");

printf("D");
printf("R");
int c1=c+1;
int b1=b-1;
m=d-b1;n=c1-a;

for(i=0;i<m;i++)
printf("D");
for(i=0;i<n;i++)
printf("L");
printf("U");
return 0;
}