PAT A Shortest Distance (20)

最新推荐文章于 2025-07-20 13:34:31 发布

xyzchenzd

最新推荐文章于 2025-07-20 13:34:31 发布

阅读量425

点赞数

CC 4.0 BY-SA版权

分类专栏： PAT 文章标签： PAT C++

本文链接：https://blog.youkuaiyun.com/xyzchenzd/article/details/26145955

PAT 专栏收录该内容

99 篇文章

订阅专栏

本文介绍了一种计算高速公路简单环形上任意两出口间最短距离的算法。通过记录各点到起点的距离，利用环形特性计算任意两点间的最小距离。输入包括出口数、相邻出口间的距离及待查询的出口对。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目

The task is really simple: given N exits on a highway which forms a simplecycle, you are supposed to tell the shortest distance between any pair ofexits.

Input Specification:

Each input file contains one test case. For each case, the first linecontains an integer N (in [3, 10⁵]), followed by N integer distancesD₁ D₂ ... D_N, where D_i is thedistance between the i-th and the (i+1)-st exits, and D_N is betweenthe N-th and the 1st exits. All the numbers in a line are separated by a space.The second line gives a positive integer M (<=10⁴), with M linesfollow, each contains a pair of exit numbers, provided that the exits arenumbered from 1 to N. It is guaranteed that the total round trip distance is nomore than 10⁷.

Output Specification:

For each test case, print your results in M lines, each contains theshortest distance between the corresponding given pair of exits.

Sample Input:

5 1 2 4 14 9

1 3

2 5

4 1

Sample Output:

保存各个点到原点的距离（不考虑环）。

任意两个点间的距离即为min(这个距离的差,总距离减这个距离）

代码：

#include <iostream>
#include <algorithm>
using namespace std;

int main()
{
	int n,*dis,tot_dis=0,temp;	//数量，各个点到原点的距离（不考虑环），总距离，临时数据
	int i;
	cin>>n;
	dis=new int[n];
	for(i=0;i<n;i++)	//输入距离，转换为到第一个点的总距离（不考虑环）
	{
		dis[i]=tot_dis;
		scanf("%d",&temp);
		tot_dis+=temp;
	}

	int m;
	cin>>m;
	int c1,c2,d_dis;
	for(i=0;i<m;i++)	//输入测试数据
	{
		scanf("%d",&c1);
		scanf("%d",&c2);
		if(c1>c2)
			swap(c1,c2);
		d_dis=dis[c2-1]-dis[c1-1];	//计算差值
		if(tot_dis-d_dis<d_dis)	//考虑环线
			d_dis=tot_dis-d_dis;
		printf("%d\n",d_dis);
	}

	delete [] dis;

	return 0;
}