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 题目：给一个环路图，算某亮点之间的最短路径长度。 解法&&思路： 普通办法会超时，采用累计长度数组a_d[MAXN]，到时候a_d[des]-a_d[source]就可以得出正向长度，注意：a_d[0] = 0; Code： #include <iostream> #include <vector> #include <algorithm> #include <string> #include <stdio.h> #include <math.h> #define MAXN 100004 int main() { int d[MAXN]; int accumulate_d[MAXN]; int n; std::cin >> n; // std::fill(accumulate_d, accumulate_d+n, 0); int total_cost = 0; for (int i = 0; i < n-1; i++) { scanf("%d", &d[i]); total_cost += d[i]; accumulate_d[i+1] = total_cost; } accumulate_d[0] = 0; scanf("%d", &d[n-1]); total_cost += d[n - 1]; int m; std::cin >> m; while (m--) { int start, des; scanf("%d %d", &start, &des); start--; des--; int normal_cost = 0; if (start > des) { int temp = start; start = des; des = temp; } int temp_cost = accumulate_d[des] - accumulate_d[start]; printf("%d\n", temp_cost>(total_cost/2)?total_cost-temp_cost:temp_cost); } system("pause"); }