1046 Shortest Distance (20分)思路分析+测试点分析

题目

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 [3,10​5​​]), followed by N integer distances D​1​​ D​2​​ ⋯ 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 (≤10​4​​), 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 10​7​​.
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

题目大意

这项任务非常简单：假设高速公路上有N个出口，形成一个简单的循环，你应该说出任何出口之间的最短距离。

输入规格：

每个输入文件包含一个测试用例。对于每种情况，第一行包含一个整数N（在[3,10^5]中），后跟N个整数距离d1d2⋯dn，其中D i是第i个出口和（i+1）-st出口之间的距离，dn是第N个出口和第1个出口之间的距离。一行中的所有数字都用空格隔开。第二行给出一个正整数M（≤10 ^4），后面是M行，每行包含一对出口编号，前提是出口编号从1到N。保证总往返距离不超过10 ^7。

输出规格：

对于每个测试用例，以M行打印结果，每行包含对应给定出口对之间的最短距离。

样本输入：

5 1 2 4 14 9

3

1 3

2 5

4 1

样本输出：

3

10

7

分析

模拟类型的题目
实际上就是求，一个点到另一个点的最短距离，是直接到这个点，还是通过环的另一半距离更短。
如果暴力解会超时，最后一个测试点通不过，所以要对数据进行预处理。这个预处理让我想起了树状数组。

代码

#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
	int n, sum = 0;
	scanf("%d", &n);
	int d[100010],dis[100010];
	for (int i = 1; i <= n; i++)
	{
		scanf("%d", &d[i]);
		sum += d[i];
		dis[i] = sum;
	}
	int m;
	scanf("%d", &m);
	for (int i = 0; i < m; i++)
	{
		int sum1 = 0;
		int a, b;
		scanf("%d %d", &a, &b);
		if (a < b) swap(a, b);
		sum1 = dis[a-1] - dis[b-1];
		printf("%d\n", min(sum1, sum - sum1));
	}
	return 0;
}

总结

我一直在纠结题目中的Di,Dn表示的距离，后来才发现i+1个点没有，而且是环形。。。被自己蠢哭了哭唧唧
注意此处预处理的方式！