1046. Shortest Distance (20)

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本文介绍了一个简单的高速公路出口间最短距离计算问题，并提供了一段C++代码实现。该算法通过预处理各出口间的距离，快速计算任意两出口间的最短路径。

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

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <algorithm>

using namespace std;

const int MAXN=100005;
int dis[MAXN]={0},A[MAXN]={0};//dis记录从1到i的距离，A记录i到i+1的距离
int sum=0;
int main()
{
   // freopen("in.txt","r",stdin);

    int N,M;

    scanf("%d",&N);
    for(int i=1;i<=N;i++)
    {
       int t;
       scanf("%d",&t);
       A[i]=t;
       sum+=t;
       dis[i]=sum;
    }

    scanf("%d",&M);
    while(M--)
    {
        int l,r;
        scanf("%d %d",&l,&r);
        if(l>r)
            swap(l,r);
        int t2=dis[r-1]-dis[l-1];
        printf("%d\n",min(t2,sum-t2));
    }


    return 0;
}