1046 Shortest Distance (20 分)

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本文介绍了一个简单的算法问题：给定形成简单环形高速公路的多个出口，如何计算任意两个出口之间的最短距离。输入包括出口数量、各段间的距离以及需要查询的出口对；输出为每对出口间的最短距离。

在这里插入图片描述
1046 Shortest Distance (20 分)
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 [3,10
5
]), followed by N integer distances D
1

D
2

⋯ D
N

, where D
i

is the distance between the i-th and the (i+1)-st exits, and D
N

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 (≤10
4
), 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 10
7
.

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

#include<iostream>
using namespace std;
int main() {
	long N;
	int M;
	cin >> N;
	int D[100005] = { 0 };
	int CycleDistance = 0;
	for (int i = 1; i <= N; i++) { cin >> D[i]; CycleDistance = CycleDistance + D[i]; D[i] = CycleDistance; }
	cin >> M;
	while (M--) {
		int x, y;
		int Distance = 0;
		cin >> x >> y;
		if (x > y) swap(x, y);
		Distance = D[y - 1] - D[x - 1];//避免重复计算
		if (Distance > CycleDistance - Distance) Distance = CycleDistance - Distance;
		cout << Distance << endl;
	}
	return 0;
}