AtCoder 2282 C - Back and Forth（模拟路径）

题目链接：http://abc051.contest.atcoder.jp/tasks/abc051_c?lang=en

C - Back and Forth

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Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

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≤1000sx,sy,tx and ty are integers.

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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: UpD: DownL: LeftR: Right

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

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Sample Input 1

Copy 0 0 1 2

Sample Output 1

Copy 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)

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Sample Input 2

Copy -2 -2 1 1

Sample Output 2

Copy UURRURRDDDLLDLLULUUURRURRDDDLLDL

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Submit

题意：二维坐标，初始坐标(sx, sy)到目的坐标(tx, ty)来回两趟的最小路径，如果有多条输出任意一条即可

解析：借用泰神的图，懒得画了，如下图

泰神博客:http://www.wonter.net/

代码：

#include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #include<vector> #include<queue> #include<map> #include<cmath> #include<string> #define N 6009 using namespace std; const int INF = 0x3f3f3f3f; typedef long long LL; int main() { int sx, sy, tx, ty; scanf("%d%d%d%d", &sx, &sy, &tx, &ty); for(int i = sy + 1; i <= ty; i++) printf("U"); for(int i = sx + 1; i <= tx; i++) printf("R"); for(int i = ty - 1; i >= sy; i--) printf("D"); for(int i = tx - 1; i >= sx; i--) printf("L"); printf("L"); for(int i = sy; i <= ty; i++) printf("U"); for(int i = sx; i <= tx; i++) printf("R"); printf("DR"); for(int i = ty; i >= sy; i--) printf("D"); for(int i = tx; i >= sx; i--) printf("L"); printf("U\n"); return 0; }