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 14 931 32 54 1

Sample Output:

3107

#include <iostream>#include <vector>using namespace std;inline int getdist(const int* const dist,int a,int b,int size){int res(0);for(int i = a;i != b;i = (i + 1) % size)res += dist[i];return res;}int main(){freopen("input.txt","r",stdin);int n;scanf("%d",&n);int* distance = new int[n];int* innerdist = new int[n + 1];innerdist[0] = 0;int total(0);for(int i = 0;i < n;++ i)scanf("%d",distance + i);for(int i = 0;i < n;++ i){total += distance[i];innerdist[i + 1] = total;}int k;scanf("%d",&k);for (int i = 0;i < k;++ i){int x,y;scanf("%d %d",&x,&y);--x,--y;int max = x > y ? x : y;int min = x > y ? y : x;int a = innerdist[max] - innerdist[min];int b = total - a;int res = a < b ? a : b;cout << res << "\n";}delete[] distance;}