PAT-A-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 [3,105]), 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:

1
2
3

3
10
7

Code

#include <iostream>
#include <algorithm>
using namespace std;

int main(){
    int n;
    scanf("%d",&n);
    int *dis = new int[n+1];//存放1号顶点到i号顶点的距离
    int *A = new int[n+1];//存放输入的边
    int sum;//存储总的距离

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

    int ans;
    scanf("%d",&ans);
    int v1,v2,distance;
    while(ans--){
        //输入顶点v1和V2
        scanf("%d%d",&v1,&v2);
        if(v1>v2)swap(v1,v2);
        int temp = dis[v2-1]-dis[v1-1];
        printf("%d\n",min(temp,sum-temp));
    }

	return 0;
}