1046 Shortest Distance（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
解题思路：
计算环形链上任意两点间的最短距离，也就是正着走和反着走哪个近。
用total记录环形的总距离，ans记录正着走的距离，输出其中较小的。
用dis[i]记录从第0个到i的距离，dis[ed-1]-dis[st-1]就是st到ed的正向距离。
dis数组在读入时就记录，如果在每一次查询中再去跑一边st到ed的距离，会超时。
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=100010;
int main(){
int n,m,total=0;
int a[maxn],dis[maxn];
dis[0]=0;
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
total+=a[i];
dis[i]=total;
}
scanf("%d",&m);
for(int i=0;i<m;i++) {
int st,ed,ans=0;
scanf("%d%d",&st,&ed);
if(st>ed){
swap(st,ed);
}
ans=dis[ed-1]-dis[st-1];
printf("%d\n",min(ans,total-ans));
}
return 0;
}