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.
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.
For each test case, print your results in M lines, each contains the shortest distance between the corresponding given pair of exits.
最后一个点运行超时：
#include <iostream> using namespace std; const int maxn = 100010; int n, m; long long dis[maxn]; long long sum_dis = 0; int e1, e2; int main() { scanf("%d", &n); for(int i = 1; i <= n; i++) { scanf("%lld", &dis[i]); //i到i+1的距离 sum_dis += dis[i]; } scanf("%d", &m); for(int i= 0; i < m; i++) { long long sum = 0; scanf("%d%d", &e1,&e2); if (e1 > e2) { int tmp = e1; e1 = e2; e2 = tmp; } for(int j = e1; j < e2; j++) { sum += dis[j]; } if (sum < sum_dis - sum) { printf("%lld\n", sum); }else { printf("%lld\n", sum_dis - sum); } } return 0; }正确code：
降低复杂度：
求出所以从1到其他结点的距离，A到B的距离（B>A）记作（B到1的距离-A到1的距离），比较环形从两边出发的长度，取最小值
#include <iostream> #include <cmath> using namespace std; const int maxn = 100010; int n, m; long long dis[maxn], a[maxn]; long long sum_dis = 0; int e1, e2; int main() { scanf("%d", &n); for(int i = 1; i <= n; i++) { scanf("%lld", &dis[i]); //i到i+1的距离 sum_dis += dis[i]; a[i] = sum_dis; } scanf("%d", &m); for(int i= 0; i < m; i++) { scanf("%d%d", &e1,&e2); if (e1 > e2) { int tmp = e1; e1 = e2; e2 = tmp; } long long temp = a[e2-1] - a[e1-1]; printf("%lld\n", min(temp, sum_dis - temp)); } return 0; }
