1046. Shortest Distance (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 14 931 32 54 1

Sample Output:

3107

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提交代码

题目大意：
有构成环的n个点，给定每个点之间的距离，然后给出m组查询，查询给出的两个点之间的最短距离。
解题思路：
不管查询哪个点都以第一个点为起始点，用一个数组记录每个点到第一个点的距离，然后要查询的两个点减去相同的距离，就是他们之间的距离。查询的时候
需要注意判断是左边的大还是右边的大，用这种方法一定是大的减去小的。但是因为成环，所以有两条不同的方向可以到达，所以我们要取两者的最小值，那
么我们就可以直接把这个环的总长度记录下来，用总长度减去一个方向的就可以得到另一个方向的了。相减的时候需要注意，是dis[r-1]-dis[l-1]。自己
画个图就很容易理解。

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;#define N 100005int dis[N];int main(){//freopen("input.txt","r",stdin);int sum = 0;int l,r,ncase,n,num;scanf("%d",&n);for(int i = 1;i <= n;i++){scanf("%d",&num);sum += num;dis[i] = sum;}scanf("%d",&ncase);while(ncase--){scanf("%d%d",&l,&r);if(l>r){swap(l,r);}int temp = dis[r-1]-dis[l-1];printf("%d\n",min(temp,sum-temp));}}