PAT 甲级 1046.Shortest Distance C++/Java

题目来源

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]), followed by N integer distances D​1​​ D​2​​ ⋯ D​N​​, where D​i​​ is the distance between the i-th and the (-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 (≤), 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 1.

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

题目大意:

输入一个N:表示有N个出口

输入N个数:D1,D2,....DN,表示第 i 个 与 第 i + 1 个出口之间的距离

输入一个M:表示有M组数据

剩下M行,每行输入2个数 i 和 j:表示 i 与 j 之间的最短距离

 

分析:

 Dn[i] :保存第 i 个与第 i + 1 个出口之间的距离

 sum :保存总距离(从1到N再到1,形成的一个环)

 

给出的两个出口: exit1, exit2 ,计算 两个出口之间的距离 s  ,再与 sum - s 比较,最小的那个就是最短距离

一开始想着用累加的方式,用D1,D2,DN计算出口之间的距离,但是这样会超时

 

所以需要对距离进行预处理,得到一个 数组 disc 

里面保存了从 1 到 2, 从 1 到 3, 从 1 到 4 的距离

如果要计算从 2 到 4 的距离, 就相当于是 1到4的距离 - 1到2的距离 

 

需要注意的是:第一个出口必须小于第二个出口,否则就交换两个值

 

C++实现:

#include <iostream>
#include <vector>
using namespace std;
int main() {
    int N;
    cin >> N;
    vector<int> Dn(N + 1);
    vector<int> disc(N + 1);
    int sum = 0;
    for (int i = 1; i <= N; ++i) {
        cin >> Dn[i];
        sum += Dn[i];
        disc[i] = sum;
    }
    int M;
    int exit1 = 0, exit2 = 0;
    cin >> M;
    for (int i = 0; i < M; ++i) {
        cin >> exit1 >> exit2;
        if (exit1 > exit2) {
            swap(exit1, exit2);
        }
        int shortest = 0;
        int temp = disc[exit2 - 1] - disc[exit1 - 1];
        shortest = temp < sum - temp ? temp : sum - temp;
        cout << shortest << endl;
    }
    return 0;
}

 

 

Java实现:

 

 

 




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