A1046.Shortest Distance

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.

输入格式：
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 D1 D2 … DN, where Di is the distance between the i-th and the (i+1)-st exits, and DN 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.
[简要翻译：每个输入包含一个测试样例。对于每一个测试样例，第一行包含一个整数N，然后跟着N个距离整数D1 D2 …….DN，Di表示第i个节点和第i+1个节点的距离，以此类推形成循环距离。第一行的每一个数用空格间隔开。第二行给出一个表示测试样例数量的正整数M，之后跟着M行，每行包含两个整数（1~N）用于测试求解。应该要保证总的往返距离不大于10^7]

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

输入样例：
5 1 2 4 14 9
3
1 3
2 5
4 1

输出样例：
3
10
7

题意：
有N个结点围成一个圈，相邻两个点之间的距离已知，且每次只能移动到相邻结点。然后给出M个询问，每个询问给出两个数字A和B即结点编号（1<=A,B<=N），求从A号结点到B号结点的最短距离。

样例解释：

思路：

注意点：

代码：

#include <iostream>  
#include <fstream>  

using namespace std;  

ifstream fin("in.txt");  
#define cin fin  

int main()  
{  
    int n;  
    cin>>n;  
    int* dis = new int[n+1];  
    int d;  
    int i;  
    int sum = 0;  
    for(i=0;i<n;i++)  
    {  
        cin>>d;  
        dis[i+1] = sum;  
        sum = sum + d;  
    }  

    int m;  
    cin>>m;  
    int begin,end;  
    int total;  
    for(i=0;i<m;i++)  
    {  
        cin>>begin>>end;  
        if(begin>end)  
        {  
            total = dis[begin]-dis[end];  
        }else  
        {  
            total = dis[end]-dis[begin];  
        }  

        if(sum-total > total)  
        {  
            cout<<total<<endl;  
        }else  
        {  
            cout<<sum-total<<endl;  
        }  

    }  
    system("PAUSE");  
    return 0;  
}