1046. Shortest Distance (20)

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本博客介绍了一个简单的任务：在形成环状的高速公路上，给出N个出口，计算任意两个出口间的最短距离。通过输入N个出口与下一个出口的距离，以及M组出口对，本算法有效计算了每组出口对之间的最短距离。

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 9
3
1 3
2 5
4 1

Sample Output:

3
10
7
在读入每个出口与下个出口的距离的时候，也要对这些距离进行加和，从而可以减少计算两个特定的出口之间的最短距离的计算量。达到一劳永逸的效果。否则，当测试点很多的时候，运行时间可能达不到要求，最后一个测试点通不过。
#include <iostream>
#include <vector>
using namespace std;
int main()
{
    vector<int> dis;
    int num_exit;
    cin >>num_exit;
    dis.push_back(0);
    int total_dis=0;
    for(int i=1;i<num_exit+1;i++)
    {
        int tmp;
        cin >>tmp;
        total_dis+=tmp;
        dis.push_back(total_dis);
    }
    int n;
    cin >>n;
    for(int i=0;i<n;i++)
    {
        int st,en;
        cin >>st >>en;
        int a,b;
        a=max(st,en);
        b=min(st,en);
        int dis1,dis2;
        dis1=dis.at(a-1)-dis.at(b-1);
        dis2=dis.back()-dis1;
        if(dis1<dis2)
            cout <<dis1 <<endl;
        else
            cout <<dis2 <<endl;
    }
    return 0;
}