a1046 shortest distance

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.

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​​.

For each test case, print your results in M lines, each contains the shortest distance between the corresponding given pair of exits.

5 1 2 4 14 9
3
1 3
2 5
4 1

3
10
7

#include <bits/stdc++.h>
using namespace std;

const int MAXN = 100001;

//cur数组为i到i+1的距离
int dis[MAXN], cur[MAXN], sum = 0;

int main()
{
    int N;
    scanf("%d", &N);
    for (int i = 1; i <= N; i++)
    {
        //输入i到i+1的距离
        scanf("%d", &cur[i]);
        //对dis数组进行预处理,以防超时
        sum += cur[i];
        dis[i] = sum;
    }

    int M, vertexA, vertexB;
    scanf("%d", &M);
    //查找最短路径
    for (int i = 1; i <= M; i++)
    {
        scanf("%d%d", &vertexA, &vertexB);
        //vertexA > vertexB,交换
        if (vertexA > vertexB)
            swap(vertexA, vertexB);
        //vertexA到vertexB的距离
        int temp = dis[vertexB - 1] - dis[vertexA - 1];
        printf("%dn", min(temp, sum - temp));
    }

    return 0;
}