PAT A1046 Shortest Distance

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本文介绍了一种算法，用于解决给定高速公路形成简单环路的N个出口间的最短距离问题。输入包括N个出口及相邻出口间的距离，通过预处理计算各出口到起点的距离，进而快速求出任意两个出口间的最短距离。

题目描述

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

Input Specification

Each input file contains one test case. For each case, the first line contains an integer N (in [3,105]), followed by N integer distances D1 D2 ⋯ DN, where Di is the distance between the i-th and the (i+1)-st exits, and DN is between the 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 lines follow, each contains a pair of exit numbers, provided that the exits are numbered from 1 to N. It is guaranteed that the total round trip distance is no more than 107.

Output Specification

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

Sample Input

5 1 2 4 14 9
3
1 3
2 5
4 1

Sample Output

3
10
7

参考代码

#include <iostream>
#include <vector>

int main() {
	int n = 0, m = 0, a = 0, b = 0, temp = 0;

	std::cin >> n;

	std::vector<int> distance(n + 1, 0);
	for (int i = 1; i < n + 1; ++i) {
		std::cin >> distance[i];
		distance[i] += distance[i - 1];
	}

	std::cin >> m;
	for (int i = 0; i < m; ++i) {
		if (i != 0) {
			std::cout << std::endl;
		}
		std::cin >> a >> b;
		temp = distance[a - 1] - distance[b - 1];
		temp = (temp < 0 ? -1 * temp : temp);
		std::cout << (temp < distance[n] - temp ? temp : distance[n] - temp);
	}

	return 0;
}

注意事项

计算两点之间距离的思路：
1. 记每个点为 Pi 按照 i 递增的顺序为正；
2. 计算每个点 Pi 到起始点 P0 正方向的距离 Di；
3. 任意两点之间的距离就等于 Di - Dj 的绝对值；
4. Pi, Pj 两点之间的另一条路径（反方向）的距离为总的距离 - Pi, Pj 两点之间的正向距离；
5. Pi, Pj 两点之间的最短距离就是正反方向距离的最小值。
distance[] 中的最后一个元素记录的是整个环路的路径长度。

题目链接

PAT A1046 Shortest Distance