1046. Shortest Distance (20) PAT 甲级

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本文介绍了一个简单的算法问题：给定一条形成简单环形的高速公路及其上的N个出口，如何计算任意两个出口之间的最短距离。输入包括出口数量、各出口间的距离及需要计算的出口对；输出则是每对出口间的最短距离。

Problem Description

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 … 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.

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

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#include<iostream>
#include<algorithm>

using namespace std;

#define MAX_N 100005


int dis[MAX_N];
int A[MAX_N];


int main(){
    int sum=0;

    int N;
    int M;
    int left,right;
    int temp;
    cin>>N;

    for(int i=1;i<=N;i++){
        cin>>A[i];
        sum+=A[i];
        dis[i]=sum;
    }
    cin>>M;
    for(int i=0;i<M;i++){
        cin>>left>>right;
        if(left>right){
            temp=left;
            left=right;
            right=temp;
        }

        temp=dis[right-1]-dis[left-1];
        cout<<min(temp,sum-temp)<<endl;
    }


}