1046 Shortest Distance (20 分)

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:

3 10 7

题目大意：

         给出n条路径长度，每条路径长度是第i个结点通往第i+ 1个结点的路径长度， 即会形成一个环，要求给出m个起点和终点求得最短距离。

 

思路：

     Dis数组存储从1号结点到下标+1号结点的距离，如dis【1】存储的是1 到2号结点的距离。每次给出的询问，将起始于结束结点的位置调整好，输出min(dis[ed – 1] – dis[st – 1], dis[n] – dis[ed – 1] + dis[st – 1])，即顺时针的路径长度（从编号小的到编号大的如1 –> 3）与其逆时针长度（逆着走）的最小值。

    

参考代码：

#include<vector> #include<cstdio> #include<algorithm> using namespace std; int n, m; vector<int> dis; int main(){ scanf("%d", &n); dis.resize(n + 1); for(int i = 1; i <= n; ++i){ scanf("%d", &dis[i]); dis[i] += dis[i - 1]; } scanf("%d", &m); for(int i = 0; i < m; ++i){ int st, ed; scanf("%d%d", &st, &ed); if(st > ed) swap(st, ed); int temp = dis[ed - 1] - dis[st - 1]; printf("%d\n", min(temp, dis[n] - temp)); } return 0; }