1.给出一个数组，求出0到i点的最短距离，可以直接从0->1->2->…->i，也可以0->n-1->n-2->n-3->…->i+1->i。
2.建立一个数组dp[i]，记录0到i之间的距离和。
3.定义sum，记录全程的路程。
4.求i，j的最短距离时，利用ans=dp[j]-[i]+num[i]-num[j]，ans2=sum-ans,min(ans,ans2)即为答案。
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 … 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 (<=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
AC代码：
[c language=”++”]
#include <iostream>
#include <stdio.h>
#include <vector>
#include <stack>
#include <algorithm>
#include <memory.h>
#include <map>
#include <set>
#include "limits.h"
using namespace std;
int main(void)
{
int n;
scanf("%d",&n);
int *num=new int[n];
int *dp=new int[n];
memset(dp, 0, sizeof(dp));
int sum=0;
for(int i=0;i<n;i++)
{
scanf("%d",&num[i]);
if(i==0) dp[i]=num[i];
else dp[i]=dp[i-1]+num[i];
sum+=num[i];
}
int m;
scanf("%d",&m);
for(int i=0;i<m;i++)
{
int a,b;
scanf("%d %d",&a,&b);
if(a>b) swap(a,b);
int tmp=dp[b-1]-dp[a-1]+num[a-1]-num[b-1];
int tmp2=sum-tmp;
printf("%d\n",min(tmp2,tmp));
}
return 0;
}
[/c]