1046. Shortest Distance (20)

时间限制：100ms
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

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此题原以为很简单，刚想疑问为啥PAT上正确率就0.2几….自己做了才发现原来会超时，因为数据量很大。
最初的想法超级简单，就是记录下每个顶点间的距离，等到有查询的时候再一个个计算~结果尼玛第三个测试点妥妥的超时了！
肯定是算法有问题，因为算法实在太简单了。
后来嘛，后来去网上一搜，搜了一半想到两个问题：
1.整个圆圈的总长度知道了，我算出一边，那么另一边的长度一减就出来了
2.从点x到点y（假设x小于y）的长度，不就是从 1到y的长度减去从1到x的长度（1点为第一个点）
而从1到i的长度，其实在读入数据的时候不就可以计算了!!!!!

dist_sum[i] = dist[i-1] + currDist;

之所以这两种做法时间上差别那么多，
因为幼稚的方法里面，每次计算时间复杂度是O(n)，一共M次查询就是O(M*n)
现在每次查询时间复杂度是O(1)，M次查询也就O(M)
题目给的规模，n=100000, M=10000
差距可想而知~
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#include
int getShortestDist(const int* dists, int N, int from, int to, long totalDist)
{
int dist_L, dist_R;
if(from > to)
{
int temp = from;
from = to;
to = temp;
}
dist_L = dists[to-1] - dists[from-1];
dist_R = totalDist - dist_L;
return (dist_L

测试点	结果	用时(ms)	内存(kB)	得分/满分
0	答案正确	0	780	14/14
1	答案正确	0	790	3/3
2	答案正确	10	1000	3/3