PAT A1046 Shortest Distance

PAT A1046 Shortest Distance

题目描述：

　　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

 

参考代码：

/***********************************************
PAT A1046 Shortest Distance
***********************************************/
#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

int main() {
    int n = 0, m = 0, sum = 0;
    
    cin >> n;

    vector<int> Distance(n + 1, 0);

    int temp = 0;
    for (int i = 1; i <= n; ++i) {
        cin >> temp;
        sum += temp;
        Distance[i] = sum;    //从0到i的距离
    }

    cin >> m;
    for (int i = 0; i < m; ++i) {
        int beg = 0, end = 0, shortDist = 0;

        cin >> beg >> end;

        if (beg > end) swap(beg, end);

        shortDist = Distance[end - 1] - Distance[beg - 1];

        cout << (shortDist < Distance[n] - shortDist ? shortDist : Distance[n] - shortDist);

        if (i != m - 1) cout << endl;
    }

    return 0;
}

 

注意事项：

　　1：提前计算好从起始点到每个点之间的距离有利于后续的计算。