pat Shortest Distance (20)

标签：sed   guarantee   namespace   names   pos   cond   ++   each   ...   

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.

输入描述：

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 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 (<=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.

输出描述：

For each test case, print your results in M lines, each contains the shortest distance between the corresponding given pair of exits.

输入例子：

5 1 2 4 14 9
3
1 3
2 5
4 1

输出例子：

3
10
7

思路：
　　题目给了我们N个点之间的距离，这N个点构成了一个环，求给出的几组点之间的最短距离。
　　一开始我想到的是建立一个N*2的数组，每行表示一个点到上个点和下个点的距离，但后来发现由于这是一个环，所以两点之间有两种走法，只要求出其中一种的距离，即可得另
　　一个。有了这个想法，写出了很粗糙的第一版

 1 #include <iostream>
 2 using namespace std;
 3 
 4 int f(int a,int b,int array[],int sum){  //判断两点之间，哪条路径短
 5     int m,n,left=0,right=0;  //right->从较小序号点向较大序号点的距离，left->另一条路径距离
 6     if(a-1<b-1){  //防止a点的序号比b点大
 7         m=a-1;
 8         n=b-1;
 9     }else{
10         m=b-1;
11         n=a-1;
12     }
13     for(int i=m;i<n;++i)
14     {
15         right+=array[i];
16     }
17     left=sum-right;
18     return left<right?left:right;
19 }
20 
21 int main()
22 {
23     int N;
24     cin>>N;
25     int array_1[N],sum=0;  //array_1->存放每个点到下个点的距离，sum->保存距离和
26     for(int i=0;i<N;++i)
27     {
28         cin>>array_1[i];
29         sum+=array_1[i];
30     }
31     int M;  //M->共M组点
32     cin>>M;
33     int array_2[M][2];
34     for(int i=0;i<M;++i)
35     {
36         cin>>array_2[i][0]>>array_2[i][1];
37     }
38     for(int i=0;i<M;++i)
39     {
40         cout<<f(array_2[i][0],array_2[i][1],array_1,sum)<<endl;
41     }
42 }

pat Shortest Distance (20)

标签：sed   guarantee   namespace   names   pos   cond   ++   each   ...   

原文地址：https://www.cnblogs.com/kiritozhj/p/9998522.html