PAT(A) 1046.Shortest Distance(20)

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

Sample Output:

3
10
7


环形图中求两点间最短距离。无非两条路，取最小即可。
注：distance 内置函数？
   dis[i]表示1号结点按顺时针方向到达"i号结点顺时针方向的下一个结点"的距离

#include<cstdio>
#include <algorithm>    //用到了swap(), min()
using namespace std;

const int MAXN=100005;
int dis[MAXN],A[MAXN];


int main()
{
    int n,left,right,query;int sum=0;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)//输入数据并记录
    {
        scanf("%d",&A[i]);//依次输入各节点间距离
        sum+=A[i];    //累加
        dis[i]=sum;   //累加一次得到一个 两节点间距离
    }
    printf("请输入你的查询次数");
    scanf("%d",&query);
    for(int i=1;i<=query;i++)
    {
        printf("请输入查询的两个结点");
        scanf("%d%d",&left,&right);
        if(left>right)    swap(left,right);
        int temp=dis[right-1]-dis[left-1];
        printf("the shortest distance:");printf("%d
",min(temp,sum-temp));//两个方向取最小即可
    }
    return 0;
}

 运行结果: