1046. Shortest Distance (20)

题目如下：

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

题目要求找出简单环路（类似表盘）中指定两点的最短距离，实质就是找出顺时针和逆时针两种路径中较小的，由于题目给出的数组规模比较庞大，因此不能简单的计算顺时针和逆时针的距离之和然后比较输出，而应当仔细思考问题的规律，试图不经过环的遍历就找到答案。

首先，我们可以在输入时累计相邻点的距离，得到环的总长sum，从而只需要计算一个方向的距离，另一个方向一减便知。

其次，我们可以再用一个数组或vector记录从起点1顺时针走到当前结点的距离，在输入时就可以得到。

最后，有了这两个量，我们可以用较大结end点到1的距离减去较小结点start到1的距离，从而得到start到end的顺时针距离，逆时针距离就等于sum-顺时针距离，从而在不遍历环的情况下得到两个距离，这样才能避免最后一个测试点超时。

通过这个题目，我反思到对于简单的题目并不能掉以轻心，应当力图寻求事物的规律来替代繁琐的过程。

#include <iostream>
#include <vector>
#include <stdio.h>
#include <math.h>

using namespace std;

vector<int> dis;
vector<int> disToOrigin;
int sum = 0;

int computeMinDis(int s, int e){
    int minDist1 = 0,minDist2 = 0;
    int start, end;
    if(s < e) { start = s; end = e; }
    else if( s == e ) return 0;
    else { start = e; end = s; }
    minDist1 = disToOrigin[end] - disToOrigin[start];
    minDist2 = sum - minDist1;
    return minDist1 < minDist2 ? minDist1 : minDist2;
}

int main()
{
    int N,M;
    cin >> N;
    dis.resize(N+1);
    disToOrigin.resize(N+1);
    disToOrigin[1] = 0;
    int value;
    for(int i = 1; i <= N; i++){
        scanf("%d",&value);
        dis[i] = value;
        if(i > 1) disToOrigin[i] = sum;
        sum += value;
    }
    cin >> M;
    int x,y;
    for(int i = 0; i < M; i++){
        scanf("%d%d",&x,&y);
        printf("%d\n",computeMinDis(x,y));
    }
    return 0;
}