【PAT（甲级）】1046 Shortest Distance（距离分析）

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​ ⋯ DN​, where Di​ is the distance between the i-th and the (i+1)-st exits, and DN​ 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

解题思路：

像这种从一个点到另一个点连续给出路程的题目，节约时间都是用sum把距离加起来的。读取数据的时候就是直接把距离加起来，然后求两点距离就是目的位置减去初始位置。

但是这题稍微复杂一点，他的路径是一个圈，所以还可以逆方向走一遍，要求找出最短的路。

所以如上图所示，如果是点1到点4的距离，就应该是7，或者30-7；点2到点5的距离就应该是21，或者31-21。也就是说，假设其实点位start，末点为end，顺时针的距离就是 dis[end-1]-dis[start-1];

逆时针的距离就是dis[N]+dis[start-1]-dis[end-1]。因为存储问题，下标的话也许会不同。

代码：

#include
using namespace std;
int N;
int dis[99999999]={0};
int rdis[99999999]={0};
 
int main(){
	scanf("%d",&N);
	for(int i=1;i<=N;i++){
		int n;
		scanf("%d",&n);
		dis[i] = n+dis[i-1];
	}
	int M;
	scanf("%d",&M);
	for(int i=0;i
		int start,end;
		scanf("%d%d",&start,&end);
		if(end
			int temp;
			temp = end;
			end = start;
			start = temp;
		}
		int min_d;
		min_d = dis[end-1]-dis[start-1];//顺时针的距离 
		
		printf("%d\n",min(min_d,dis[N]+dis[start-1]-dis[end-1]));//比较顺时针与逆时针的距离谁小 
	}
	return 0;
}