PAT 1046 Shortest Distance (20分)

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最新推荐文章于 2025-03-14 11:15:13 发布

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分类专栏： c/c++ algorithm PAT 文章标签：算法 c++

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本文介绍了一种计算高速公路形成简单环上任意两个出口间最短距离的算法。通过预处理每个节点到起始节点的距离，算法能够快速确定两节点间的劣弧或优弧距离，从而找出最短路径。

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题目

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

解析

题目：

n个节点形成一个圆环，编号从1到n，按顺序给出 1-2距离、2-3距离、…、n-1距离。给出m对节点i,j，要求输出i，j之间的最短距离。

思路：

在获取输入的同时计算出每个结点到第一个结点的距离，并将它存放在数组dis[]中，即dis[n]表示节点n到节点1的顺序方向对应的距离，dis[n+1]表示整个环的距离和。
任意两点的距离要么环的"劣弧"，要么是环的"优弧"，这里先求按顺序由a到b的距离（a<b）,也就是dis[b]-dis[a]，另一端距离用圆环总长度（dis[n+1]）减去这里求的距离，比较两者取最小值。
圆环总长度小于10⁷，那么距离值可以定义为int.

代码

#include <iostream>
#include <vector>
using namespace std;

int main() {
    int n, dis[100001], temp;
    cin >> n;
    // 节点编号从1开始，
    // 输入数据是 1-2距离，2-3距离，...n-1距离
    for (int i = 1; i <= n; ++i) {
        cin >> temp;
        // dis[i]保存从节点1到节点i的距离，假如有n和节点，那么dis[n + 1] 就是整个环的距离和
        if (i == 1) dis[i + 1] = temp;
        else dis[i + 1] = dis[i] + temp;
    }
    int m, a, b;
    cin >> m;
    while (m-- > 0) {
        cin >> a >> b;
        // 保证按顺序小，大
        if (a > b) swap(a, b);
        // “劣弧”对应的长度
        temp = dis[b] - dis[a];
        // dis[n + 1] - temp 是 “优弧”对应的长度，取较小的那个
        cout << min(temp, dis[n + 1] - temp) << endl;
    }

    return 0;
}