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 ... 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
#include<cstdio> #include<algorithm> using namespace std; const int maxn = 100010; int dis[maxn],A[maxn]; int main(){ int sum = 0,n,left,right; scanf("%d",&n); for(int i = 1; i <= n; i++){ scanf("%d",&A[i]); sum += A[i]; dis[i] = sum; // dis代表从1个顶点到第i个顶点的距离 } int m; scanf("%d",&m); for(int i = 0; i < m; i++){ scanf("%d%d",&left,&right); if(right < left) swap(left,right); int temp = dis[right - 1] - dis[left - 1]; printf("%d\n",min(temp,sum - temp)); } return 0; }