A1046 Shortest Distance 最短路径

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]), followed by Ninteger distances D​1​​ D​2​​ ⋯ D​N​​, where D​i​​ is the distance between the i-th and the (-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(≤), 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 1.

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：

•数据构成环，使用dis[i]存储i+1点到第1个点的距离，比较dis[right-1]-dis[left-1]和sum-(dis[right-1]-dis[left-1])得到shortest distance，此时时间复杂度是O(1)；

•不存在顺序问题，当左大右小时，swap(left,right);

•数组存储时将第一个点到第一个点存在了dis[0],转一圈的距离存到dis[n];

以上参考《算法笔记》和我自己的理解(*/ω＼*)

#include <iostream>
#include <algorithm> //min swap
#include <cstring>
using namespace std;
const int MAX = 100005;
int main() {
    int n, num, m, left, right;
    int dis[MAX];
    memset(dis, 0, sizeof(dis));
    cin >> n;
    for (int i = 1; i <=n; i++) {
        cin >> num;
        dis[i] = dis[i - 1] + num;
    }
    int sum = dis[n];
    cin >> m;
    for (int i = 0; i < m; i++) {
        cin >> left >> right;
        if (right < left)swap(left, right);
        //int temp = dis[right - 1] - dis[left - 1];
        //cout << min(temp, sum - temp) << endl;
        cout << min(dis[right - 1] - dis[left - 1], sum - (dis[right - 1] - dis[left - 1]))<<endl;
    }
    return 0;
}