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 D₁D₂ ... 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

来源： <http://www.patest.cn/contests/pat-a-practise/1046>

 

#include <iostream>
#include <vector>
#include <stdio.h>
#pragma warning(disable:4996)
using namespace std;
vector<int> num;
int main(void) {
	int n,sum = 0;
	cin >> n;
	for (int i = 0; i < n; i++) {
		int roadtemp;
		scanf("%d", &roadtemp);
		
		sum += roadtemp;
		num.push_back(sum);
	}
	int m;
	cin >> m;
	for (int i = 0; i < m; i++) {
		int start, end, change;
	//	cin >> start >> end;
		scanf("%d %d", &start, &end);
		if (start > end) {
			change = end;
			end = start;
			start = change;
		}
		int road = 0;
		
		if (start == 1) {
			road = num[end-2];
		}
		else {
			road = num[end-2] - num[start-2];
		}
		if (road > num[n-1] - road)
			road = num[n-1] - road;
		//cout << road << endl;
		printf("%d\n", road);
	}
	return 0;
}

来自为知笔记(Wiz)