PAT 1046

1046. Shortest Distance (20)

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 ... 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 (<=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
这题做这么久真是没治了，本来不是很难的一题，被自己蠢到了……

#include <iostream>
#include <cmath>
using namespace std;

int exitPort[100005]; //开始少一个0，可见细心的重要性， 10^5并不是10005，不是五位数谢谢

int main()
{
    int N;
    cin >> N;

    exitPort[1] = 0;

    int sum = 0;
    for(int i = 2; i <= N + 1; i++)
    {
        int road;
        cin >> road;
        sum += road;
        exitPort[i] = sum;
    }

    int M;
    cin >> M;
    while(M--)
    {
        int in, out;
        cin >> in >> out;

        if(sum - abs(exitPort[out] - exitPort[in]) > abs(exitPort[out] - exitPort[in])) //由于写成了abs(sum - exitPort[out] + exitPort[in]) 有一个点一直没过
        {
            cout << abs(exitPort[out] - exitPort[in]) << endl;
        }
        else
            cout << sum - abs(exitPort[out] - exitPort[in]) << endl;
    }

    return 0;
}

 

这题唯一的收获就是定义数组的时候最好不要用exit做名称，有的OJ来说， exit是内置的变量，可能会有编译错误，因此修改为exitPort