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📚专栏地址：PAT题解集合
📝原题地址：题目详情 - 1046 Shortest Distance (pintia.cn)
🔑中文翻译：最短距离
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1046 Shortest Distance

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

题意

在一个环形的公路上有 n 个出口，现在给定一个序列 {D1 D2 ⋯ DN} ，其中 Di 表示第 i 个出口与第 i+1 个出口之间的距离， Dn 表示第 n 个出口与第 i 个出口之间的距离。

现在给定 m 个询问，需要我们输出给定两个出口之间的距离。

思路

这道题可以用前缀和的方法来做，创建一个数组 s 记录距离总和，其中 s[i] 表示第 1 个出口到第 i+1 个出口的距离之和。

这样一来，我们就可以通过 s[r-1]-s[l-1] 来计算出第 r 个出口与第 l 个出口之间的距离了。

但是，需要注意的是这是一个环形公路，所以还要计算 r...1...l 这段路径，然后与上面计算的路径作比较取最小值输出。而这段路径可以先计算出第 r 个路口到第 1 个路口的距离，即 s[n]-s[r-1] 。然后再计算第 1 个路口的距离到第 l 个路口的距离，即 s[l-1] 。两者相加就是这段距离之和，即 s[n]-s[r-1]+s[l-1] 。

故最终的答案就使求上面两个路径的最小值并输出，即 min(s[r-1]-s[l-1],s[n]-s[r-1]+s[l-1]) 。

代码

#include
using namespace std;const int N = 100010;
int s[N];
int n, m;int main()
{//输入数组元素并求前缀和scanf("%d", &n);for (int i = 1; i <= n; i++){scanf("%d", &s[i]);s[i] += s[i - 1];}//进行询问scanf("%d", &m);while (m--){int l, r;scanf("%d%d", &l, &r);if (l > r) swap(l, r);printf("%d\n", min(s[r - 1] - s[l - 1], s[n] - s[r - 1] + s[l - 1]));}return 0;
}