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 … 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
解题思路：如果只是简单的记录各个点之间的距离，每次询问的时候进行计算的话，最后一个测试点会运行超时，于是就改成数组中记录的是，从一号点到当前点的总距离，查询的时候直接将距离相减就是正向的两个点之间的距离，反向的距离用总的环的距离减去正向两个点之间的距离就可，输出较小的距离。

#include<iostream>#include<vector>#include<stdio.h>using namespace std;int main(){    for (int n; scanf("%d", &n) != EOF;){        int *arr = new int[n+1];        int total = 0;        for (int i = 1; i <= n; i++){            arr[i] = total;            int temp;            scanf("%d", &temp);            total += temp;        }        int m;        scanf("%d", &m);        while (m--){            int begin;            int end;            scanf("%d%d", &begin, &end);            if (begin > end){                begin = begin^end;                end = begin^end;                begin = begin^end;            }            int distance = arr[end] - arr[begin];            if (distance > total - distance){                printf("%d\n", total - distance);            }            else{                printf("%d\n", distance);            }        }    }}