PAT A1046 Shortest Distance

前言

传送门

正文

题目描述

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 (≤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 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

思路:

数组a[]存放相邻地点之间的距离，再对每个输入的距离进行累加即可得到整个环一周的长度，同时在输入数据的时候用dis[]数组记录下顺时针每个地点到第1个地点的距离 注: a[i]表示第i个地点与第i+1个地点之间的距离 dis[i]表示顺时针第1个地点到第i+1个地点之间的距离故两点之间距离可用diastance=dis[right-1]-dis[left-1]来表示(顺时针)，再和sum-distance比较就可以得出最短的距离

参考题解：

#include<cstdio> const int N=100010; int a[N]={0},dis[N]={0}; int main(){ int num,sum=0,m,left,right,shortest; scanf("%d",&num); //a[i]表示第i个地点与第i+1个地点之间的距离 //dis[i]表示顺时针第1个地点到第i+1个地点之间的距离 for(int i=1;i<=num;i++){ scanf("%d",&a[i]); sum+=a[i]; dis[i]=sum; } scanf("%d",&m); while(m--){ scanf("%d%d",&left,&right); if(right<left){ int temp=left; left=right; right=temp; } int distance=dis[right-1]-dis[left-1]; shortest=(distance>sum-distance)?sum-distance:distance; printf("%d\n",shortest); } return 0; }

后记

功名桥，世俗道，年少难免走一遭