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 [3,10​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​​.

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

其實題目的意思挺簡單的。。。

一開始用的鄰接表+spfa，第三組段錯誤。。不知道為啥。。

應該出題的意思不是讓你用這個方法吧。。

然後用的字首和。可以的。。（只不過一開始沒有想到。。。）

AC程式碼：

 #include <bits/stdc++.h>
///就兩條路，求字首題目，意思是陣列的某個位置上的數，
///存的是在此之前的所有的數的和。
using namespace std;
int main()
{
    int n;
    int a[100005];
    long long pre[100005];///字首。
    long long sum = 0;///總和。
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        sum += a[i];
        pre[i] += (pre[i-1] + a[i]);
    }
    int q;
    int u,v;
    scanf("%d",&q);
    while(q--)
    {
        scanf("%d%d",&u,&v);
        if(u>v)
        {
            int t = u;
            u = v;
            v = t;
        }
        long long mn = min(pre[v-1]-pre[u-1],sum-pre[v-1]+pre[u-1]);
        printf("%lld\n",mn);
    }
    return 0;
}