1046. Shortest Distance (20)

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题目链接：http://www.patest.cn/contests/pat-a-practise/1046

题目：

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:

分析：

形成一个环路的路程长度比较：

前面都累加，然后比较s到j的距离s1和s到起点和t到终点的距离和再加上终点到起点的距离和s2的大小
distance[t] - distance[s] 和 distance[s] + distance[n + 1] + distance[n] - distance[t]

AC代码：

#include<iostream>#include<stdio.h>#include<string.h>using namespace std;int main(){ //freopen("F://Temp/input.txt", "r", stdin); int n, m; int distance[100002]; cin >> n; memset(distance, 0, sizeof(distance)); for (int i = 2; i <= n; i++){  cin >> distance[i]; } for (int i = 2; i <= n; i++){  distance[i] += distance[i - 1]; } cin >> distance[n + 1];//前面都累加，但是最后一个是起点到终点的距离不用累加。 cin >> m; for (int i = 0; i < m; i++){  int s, t, min = 0x7fffffff;  cin >> s >> t;  if (s > t){   int tmp = s;   s = t;   t = tmp;  }  if (distance[t] - distance[s] < distance[s] + distance[n + 1] + distance[n] - distance[t]){   min = distance[t] - distance[s];  }  else min = distance[s] + distance[n + 1] + distance[n]- distance[t];  cout << min << endl; } return 0;}

截图：

——Apie陈小旭

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