AtCoder

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

感觉智商严重不足，因为同样的格子不能走重复，所以最短路径一定是

。。。我还傻逼似的去bfs加标记数组加记录前缀输出路径。。。最近想问题老是特别僵化，看到题目头脑根本转不起来。。。

平时练习一定要多花时间自己想题目训练思维，不然比赛时会花比别人多几倍的时间去写，因为你根本不熟悉这个思维，看题解只会增加记忆对思维的训练微乎其微，要多想

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <vector>#include <map>using namespace std;typedef long long LL;const int N = 1100000;int vis[2010][2010];int main(){    int s1, e1, s2, e2;    scanf("%d %d %d %d", &s1, &e1, &s2, &e2);    for(int i=1;i<=(e2-e1);i++) printf("U");    for(int i=1;i<=(s2-s1);i++) printf("R");    for(int i=1;i<=(e2-e1);i++) printf("D");    for(int i=1;i<=(s2-s1);i++) printf("L");    printf("L");    for(int i=1;i<=(e2-e1)+1;i++) printf("U");    for(int i=1;i<=(s2-s1)+1;i++) printf("R");    printf("D");    printf("R");    for(int i=1;i<=(e2-e1)+1;i++) printf("D");    for(int i=1;i<=(s2-s1)+1;i++) printf("L");    printf("U\n");    return 0;}