PAT-A1046 Shortest Distance

最新推荐文章于 2020-01-28 20:46:06 发布

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最新推荐文章于 2020-01-28 20:46:06 发布

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本文介绍了一个简单的编程问题：计算环形高速公路上任意两个出口之间的最短距离。输入包括出口数量、各段距离及询问对，输出每对出口间的最短路径长度。

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1046 Shortest Distance（20 分）

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 Diis 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

#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
	int n;
	int dist[100010];
	int sum = 0;
	int m;
	scanf("%d", &n);
	for(int i = 1; i <= n; i++){
		scanf("%d", &dist[i]);
		sum += dist[i];
		dist[i] = sum;  //dist[i]累加，避免之后重复计算导致超时
	} 
	scanf("%d", &m);
	int d1, d2;
	for(int i = 1; i <= m; i++){
		scanf("%d%d", &d1, &d2);
		if(d1 > d2) swap(d1,d2);
		int  ans = dist[d2-1] - dist[d1-1];
		printf("%d\n", min(ans, sum - ans));
	} 
	return 0;
}