1046 Shortest Distance (20分) 【简单模拟】

 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

解题思路：

首先，用一个数组dist2[i]保存i点到起点1的距离，这样当计算a，b两点间的距离则为tmp = |dist2[a]-dist2[b]|

此时，只要在比较一下tmp与总距离的一半sum/2之间的大小，若tmp小则输出tmp，否则输出sum-tmp。

#include<iostream>
#include<math.h>
using namespace std;

int dist1[100010], dist2[100010];   //dist1保存两点间的距离，dist2保存i到1点的距离

int main()
{
	int n;
	cin >> n;
	int sum = 0;
	for (int i = 1; i <= n; i++)
	{
		cin >> dist1[i];
		sum += dist1[i];
		dist2[i] = sum - dist1[i];
	}
	int q;
	cin >> q;
	while (q--)
	{
		int a, b;
		cin >> a >> b;
		int tmp = abs(dist2[a] - dist2[b]);
		if (tmp < sum / 2)
			cout << tmp << endl;
		else
			cout << sum - tmp << endl;
	}
	return 0;
}