PAT甲级 1046. Shortest Distance (20)

题目：

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 ... 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 (<=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

思路：

分为循环正向相加和反向相加两种选择，都进行计算，取最小值。

代码：

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

int main()
{
	//ifstream cin;
	//cin.open("case1.txt");
	

	int dis[100001] = { 0 };
	int total_dis[100001] = { 0 };
	
	int N;
	cin>>N;
	int i;
	for (i = 1; i <= N; ++i)
	{
		cin >> dis[i];
		total_dis[i] = total_dis[i - 1] + dis[i];
	}

	int M,n1,n2,tmp,d1,d2;
	cin >> M;

	vector<vector<int>> p(M,vector<int>(2));
	for (i = 0; i < M; ++i)
	{
		cin >> p[i][0] >> p[i][1];

	}

	for (i = 0; i < M; ++i)
	{
		n1 = p[i][0];
		n2 = p[i][1];
		
		if (n2 < n1)
		{
			tmp = n2;
			n2 = n1;
			n1 = tmp;
		}
		d1 = total_dis[n2 - 1] - total_dis[n1 - 1];

		d2 = total_dis[N] - total_dis[n2 - 1];
		d2 += total_dis[n1 - 1];

		if (d1 < d2)
			cout << d1;
		else
			cout << d2;
		cout << endl;
	}
	system("pause");
	return 0;
}