PAT A Shortest Distance (20)

题目

The task is really simple: given N exits on a highway which forms a simplecycle, you are supposed to tell the shortest distance between any pair ofexits.

Input Specification:

Each input file contains one test case. For each case, the first linecontains an integer N (in [3, 10⁵]), followed by N integer distancesD₁ D₂ ... D_N, where D_i is thedistance between the i-th and the (i+1)-st exits, and D_N is betweenthe 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 linesfollow, each contains a pair of exit numbers, provided that the exits arenumbered from 1 to N. It is guaranteed that the total round trip distance is nomore than 10⁷.

Output Specification:

For each test case, print your results in M lines, each contains theshortest 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

 

保存各个点到原点的距离（不考虑环）。

任意两个点间的距离即为min(这个距离的差,总距离减这个距离）

 

代码：

#include <iostream>#include <algorithm>using namespace std;int main(){int n,*dis,tot_dis=0,temp;//数量，各个点到原点的距离（不考虑环），总距离，临时数据int i;cin>>n;dis=new int[n];for(i=0;i<n;i++)//输入距离，转换为到第一个点的总距离（不考虑环）{dis[i]=tot_dis;scanf("%d",&temp);tot_dis+=temp;}int m;cin>>m;int c1,c2,d_dis;for(i=0;i<m;i++)//输入测试数据{scanf("%d",&c1);scanf("%d",&c2);if(c1>c2)swap(c1,c2);d_dis=dis[c2-1]-dis[c1-1];//计算差值if(tot_dis-d_dis<d_dis)//考虑环线d_dis=tot_dis-d_dis;printf("%d\n",d_dis);}delete [] dis;return 0;}