PAT 1046 Shortest Distance (20)（20 分）

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本文介绍了一种算法，用于计算高速公路形成简单环形时任意两个出口之间的最短距离。输入包括环形路径上的出口数量及各段距离，以及若干对出口编号以求解最短距离。文章提供了AC代码实现。

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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

解答：

（1）我刚开始做这道题时，一直没搞清题目意思，不能理解用例。后来终于理解（如图）。然后因为两点形成cycle，因此两点之间有两条路径，分别比较即可。

（2）但是如果将每条路径的长度存储下来，然后对M次查询都进行计算的话，会有一个用例超时，这时我们需要思考更加有效的方式。因此我们可以存储从路径1到路径N的总长度，待到查询时，只要相加减即可，顺利AC。

AC代码如下:

#include<iostream>
#include<cstdio>
#include<vector>

using namespace std;

int main()
{
	vector<int> ivec; int sum = 0;
	int N;
	
	scanf("%d", &N);
	while(N--)
	{
		int val;
		scanf("%d", &val); sum += val;
		ivec.push_back(sum);
	}
	int M, index1, index2;
	scanf("%d", &M); 
	for(int i = 0; i < M; ++i)
	{
		scanf("%d %d", &index1, &index2);
		if(index1 > index2) swap(index1, index2);
		int dis1 = ivec[index2-2] - (index1 == 1 ? 0 : ivec[index1 - 2]);
		int dis2 = ivec.back() - ivec[index2-2] + (index1 == 1 ? 0 : ivec[index1 - 2]);
		printf("%d\n", min(dis1, dis2));
	}
	return 0;
}