A1046 Shortest Distance (20 point(s))

简单圈内两点最短距离计算

1. 原文

原题

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 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 (≤$10^4$), 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

2. 解析

有n个结点围成一个圈，相邻两个点之间的距离已知，每次只能移动到相邻点。询问A和B之间的最短距离。

如1，2，3，4，5

1 3: 从1号到3号的最短距离为3 路径为1->2->3

2 5: 从2号到5号的最短距离为10 路径为2->1->5

4 1: 从4号到1号的最短距离为7 路径为4->3->1

Solution:

用num[i]表示1号结点按顺时针方向到达”i号结点顺时针方向的下一个结点”，sum表示一圈总距离。每个查询left->right，结果就是num(left,right) 和sum-num(left, right)之间的较小值。

3. AC代码

#include<cstdio>
int num[100010]={};
int main(){
	int n;
	scanf("%d",&n);
	int sum = 0;
	for (int i = 1; i <= n; ++i)
	{
		int a;
		scanf("%d",&a);
		sum+=a;
		num[i] = num[i-1]+a;
	}
	int m;
	scanf("%d",&m);
	for (int i = 0; i < m; ++i)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		int temp = 0;
		if (a>b)
		{
			temp = num[a-1]-num[b-1];
		}else{
			temp = num[b-1]-num[a-1];
		}
		if (sum-temp>temp)
		{
			printf("%d\n",temp);
		}else{
			printf("%d\n", sum-temp);
		}
	}
	
	return 0;
}