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]), followed by N integer 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

题意：

　　给出由数字组成的环，以及环中相邻两点间的距离，求出任意两结点间的最小距离。

思路：

　　任意两节点间的最小距离包括两种情况，即顺时针方向上有一个距离，逆时针方向上有一个距离。我们用两个数组sum1和sum2来分别存储由起点（1）开始依次向两个方向上的各个点的距离和。当求a， b两点间的最小距离的时候，我们只需要求min(sum1[a] - sum1[b], sum1[b] + sum2[a])即可（确保a > b）。

Code :

#include <bits/stdc++.h>

using namespace std;

int main() {
    int n;
    cin >> n;
    vector<int> dist(n + 1, 0);
    vector<int> sum1(n + 1, 0);
    vector<int> sum2(n + 1, 0);
    for (int i = 1; i <= n; ++i) cin >> dist[i];
    for (int i = 1; i <= n; ++i) sum1[i] = sum1[i - 1] + dist[i - 1];
    for (int i = n; i >= 1; --i) {
        if (i == n)
            sum2[i] = dist[i];
        else if (i == 1)
            sum2[i] = 0;
        else
            sum2[i] = sum2[i + 1] + dist[i];
    }
    int m, a, b;
    cin >> m;
    for (int i = 0; i < m; ++i) {
        cin >> a >> b;
        if (a < b) swap(a, b);
        int temp = min(sum1[a] - sum1[b], sum1[b] + sum2[a]);
        cout << temp << endl;
    }
    return 0;
}