1046. Shortest Distance (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

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 100004int 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");}