1046. Shortest Distance (20)

最新推荐文章于 2021-08-09 22:12:34 发布

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本文介绍了一个简单的算法，用于计算高速公路形成简单环形时任意两个出口之间的最短距离。输入包括环形高速上的出口数量及各段距离，以及需要查询距离的出口对；输出则是每对出口间的最短距离。

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
#include "iostream"
#include "algorithm"
#include "cstring"
using namespace std;
int node[100005];
int main() {
	int n;
	int total = 0;
	cin >> n;
	node[0] = 0;
	for (int i = 1; i <= n; i++) {
		scanf("%d", &node[i]);
		total += node[i];
		node[i] += node[i - 1];
	}
	int m;
	scanf("%d", &m);
	while (m--) {
		int a, b;
		int sum1 = 0;
		cin >> a >> b;
		if (a > b) {
			int t = a;
			a = b;
			b = t;
		}
		printf("%d\n", min(node[b-1]-node[a-1],total-node[b-1]+node[a-1]));
	}
	return 0;
}