PAT1046: Shortest Distance

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 9
3
1 3
2 5
4 1

Sample Output:

3
10
7

思路

1.直接用累加求和，结果最后一项测试用例超时。
2.开一个二维数组暴力枚举出所有可能的情况，检测时直接输出对应情况，结果内存超了(最坏情况数组大小100001 * 100001）。
3.保存每一个目标点到起始点的road[i]，那么任意两个点s,e的距离可以通过road[s]-road[e]计算出来，而不用再去一步步累加，所以不会超时。而且这种解决方案最坏情况开辟的数组大小为100001，不会超内存。

代码

#include<iostream>
#include<vector>
#include<math.h>
using namespace std;


int main()
{
   int N;
   while(cin >> N)
   {
       vector<int> road(N + 2,0);
       int sum = 0;
       for(int i = 1;i <= N;i++)
       {
           int value;
           cin >> value;
           road[i + 1] = road[i] + value;
           sum += value;
       }
        int M;
        cin >> M;
        while(M--)
        {
            int s,e;
            cin >> s >> e;
            int path = abs(road[s] - road[e]);
            cout << min(path,sum - path) << endl;
        }
   }
}