PAT 甲级 1046 Shortest Distance (20 分)（前缀和,想了一会儿）

1046 Shortest Distance (20 分)

 

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]), followed by N integer distances D​1​​ D​2​​ ⋯ D​N​​, where D​i​​is the distance between the i-th and the (-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 (≤), 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 1.

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

题意：

　　给出一个环形的高速公路，其中有N个出口，第Di​个出口是i到i-1的距离，而DN是N到1 的距离。给出任意两个出口，计算两者的最短距离。

题解：

显而易见，这是一个循环队列，计算距离需要考虑两个方向，但是如果直接遍历的话，时间复杂度为O(n2) O(n^2)O(n 2)这个数据量会超时（第三个测试点），所以我们需要考虑，如何优化这个距离计算过程。不妨考虑，计算每个出口两个方向的累加距离，这样计算两者之间的距离的时候，直接做加减即可，时间复杂度为O(n) O(n)O(n)。

AC代码：

#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<string>
#include<cstring>
using namespace std;
int n;
int a[100005];
int s1[100005];
int s2[100005];
int main()
{
    cin>>n;
    memset(s1,0,sizeof(s1));
    memset(s1,0,sizeof(s2));
    for(int i=1;i<=n;i++){
        cin>>a[i];
        s1[i+1]=s1[i]+a[i];
    }
    for(int i=1;i<=n;i++){
        s2[i+1]=s2[i]+a[n-i+1];
    }
    /*for(int i=1;i<=n;i++){
        cout<<i<<" s1 "<<s1[i]<<endl;
        cout<<i<<" s2 "<<s2[i]<<endl;
    }*/
    int m;
    int u,v;
    cin>>m;
    for(int i=1;i<=m;i++){
        cin>>u>>v;
        if(u>v){
            int temp=v;
            v=u;
            u=temp;
        }
        cout<<min(s1[v]-s1[u],s2[n+2-v]+s1[u])<<endl;//两个方向选最大
    }
    return 0;
}