https://pintia.cn/problem-sets/994805342720868352/problems/994805435700199424
1046 Shortest Distance (20)（20 分）
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思路
这题思路还是蛮重要的，设计一个dis【】数组记录表示点1到点i的距离，因为题目要求点u到点v的最短距离：
分两种情况：
1、求dis(u,v)就是dis[v-1]-dis[u-1];
2、因为路径本身是个环，另有：sum-dis(u,v)
求这两种情况的最小值
#include <iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int nmax=1e5+1;
const int mmax=1e4;
int a[nmax]; //a[i]表示点第i条边，点i到点i+1的距离
int dis[nmax];//dis[i]表示点1到点i的距离
int main(int argc, char** argv) {
int n;
cin>>n;
int sum=0;
for(int i=1;i<=n;i++){
cin>>a[i];
sum+=a[i];
dis[i]=sum;
}
int m;
cin>>m;
int u,v;
int dist;
for(int i=0;i<m;i++){
cin>>u>>v;
if(u>v){
swap(u,v);
}
int tmp=dis[v-1]-dis[u-1];
int ans=min(tmp,sum-tmp);
cout<<ans<<endl;
}
return 0;
}