PAT(A) 1046. Shortest Distance (20)

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 <algorithm>    //用到了swap(), min()
using namespace std;
#include <cstdio>

const int MAXN=100005;
int onetodis[MAXN], dis[MAXN];  //1号结点到i号结点顺时针方向的下一个结点的距离，i号到下一个结点的距离
//例： onetodis[5] 即 1->5->1的距离

int main()
{
    int n, sum=0, M;    //总点数，一圈的总距离，M组待查的路径
    int start, end;     //待查找距离的两点的编号
    scanf("%d", &n);
    for(int i=1; i<=n; i++){ 　　//存入1->n中
        scanf("%d", &dis[i]);   //表示i->i的下一个结点的距离
        sum += dis[i];
        onetodis[i]=sum;        //预处理onetodis[] 表示1->i的下一个结点的距离
    }
    scanf("%d", &M);    //待查的M组编号的距离
    for(int i=0; i<M; i++){
        scanf("%d%d", &start, &end);
        if(start>end) swap(start, end);     //eg. 5->2 转化为求 2->5
        int tmpdis=onetodis[end-1]-onetodis[start-1]; //start~end = (1~end)-(1->start)
        printf("%d
", min(tmpdis, sum-tmpdis));      //取两边最小的距离
    }
    return 0;
}