1046 Shortest Distance

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 [3,10^5^]), followed by N integer distances D~1~ D~2~ ⋯ 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 (≤10^4^), 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 10^7^.

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:

1
2
3

3
10
7

浅析

注意 dis 数组的使用，其表示从结点 1 顺时针到结点 n 的距离，如果只考虑了 sum，是会超时滴～

Code

#include 
#include 
#include 
using namespace std;
int main(){
    int n, sum = 0;
    cin >> n;
    vector<int> v(n+1);
    vector<int> dis(n + 1, 0);
    for (int i = 1;i<=n;i++){< span>
        cin >> v[i];
        sum += v[i];
        dis[i] = sum;
    }
    int k;
    cin >> k;
    while(k--){
        int a1, a2;
        cin >> a1 >> a2;
        if(a1>a2)
            swap(a1, a2);
        int temp = dis[a2-1] - dis[a1-1];
        temp = min(temp,sum - temp);
        printf("%d\n", temp);
    }
}