浙江大学PAT_甲级_1046. Shortest Distance (20)

题目链接：点击打开链接

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 14 931 32 54 1

Sample Output:

3107

我的C++程序：

#include<iostream>#include<algorithm>#include<vector>using namespace std;int main(){int n,m;//n个点,m对两点距离int temp,min_distance=0,cycle_length=0;//min_distance两点最短距离，cycle_length圈长度int a, b;//起点a,终点bint i = 0;int distance[100000];//istance[i]存储第i个点到第一个点的距离，i>=1cin >> n;for (i = 1;i <= n;i++){cin >> temp;cycle_length += temp;//计算圆圈长度distance[i + 1] = distance[i] + temp;}cin >> m;while (m--){cin >> a >> b;min_distance = abs(distance[b] - distance[a]);cout << min(min_distance, cycle_length- min_distance)<< endl;//圆圈内的两点最短距离}//system("pause");return 0;}