题目
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:
代码语言：javascript复制5 1 2 4 14 9
3
1 3
2 5
4 1Sample Output:
代码语言：javascript复制3
10
7解析
题目：
n个节点形成一个圆环，编号从1到n，按顺序给出 1-2距离、2-3距离、...、n-1距离。给出m对节点i,j，要求输出i，j之间的最短距离。
思路：
- 在获取输入的同时计算出每个结点到第一个结点的距离，并将它存放在数组
dis[]中，即dis[n]表示节点n到节点1的顺序方向对应的距离，dis[n 1]表示整个环的距离和。 - 任意两点的距离要么环的"劣弧"，要么是环的"优弧"，这里先求按顺序由
a到b的距离（a<b）,也就是dis[b]-dis[a]，另一端距离用圆环总长度（dis[n 1]）减去这里求的距离，比较两者取最小值。 - 圆环总长度小于107，那么距离值可以定义为int.
代码
代码语言：javascript复制#include <iostream>
#include <vector>
using namespace std;
int main() {
int n, dis[100001], temp;
cin >> n;
// 节点编号从1开始，
// 输入数据是 1-2距离，2-3距离，...n-1距离
for (int i = 1; i <= n; i) {
cin >> temp;
// dis[i]保存从节点1到节点i的距离，假如有n和节点，那么dis[n 1] 就是整个环的距离和
if (i == 1) dis[i 1] = temp;
else dis[i 1] = dis[i] temp;
}
int m, a, b;
cin >> m;
while (m-- > 0) {
cin >> a >> b;
// 保证按顺序小，大
if (a > b) swap(a, b);
// “劣弧”对应的长度
temp = dis[b] - dis[a];
// dis[n 1] - temp 是 “优弧”对应的长度，取较小的那个
cout << min(temp, dis[n 1] - temp) << endl;
}
return 0;
}
