AtCoder 2282 C - Back and Forth（模拟路径）

最新推荐文章于 2023-04-20 16:42:06 发布

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最新推荐文章于 2023-04-20 16:42:06 发布

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本文解析了一个AtCoder上的路径寻找问题，目标是最短路径地从起点到终点往返两次，同时避免重复经过同一位置（除起止点外）。文章提供了一种解决方案，并附带详细步骤说明。

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

C - Back and Forth

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≤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

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)

Sample Input 2

Copy

-2 -2 1 1

Sample Output 2

Copy

UURRURRDDDLLDLLULUUURRURRDDDLLDL

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;
}