技術標籤：資料結構_基礎

原題

點此連結1

題目分析

最短路徑問題，所求即 1.從哪個點開始 2.到最遠的點的距離 3.最短。
最一般的解法就是分別求出每個點到最遠點的距離，然後取最短。解題時需要考慮一個問題：
如何求解從點A開始的最遠點距離：在使用BFS搜尋的過程中，已訪問過的點同樣需要考慮，如果該點已訪問，但是本次訪問到這個點的路徑長度優於該結點記憶體有的最短路徑，則仍需要更新最短路徑並壓入佇列中

程式碼

#include <iostream>
#include <map>
#include <vector>
#include <queue>
 

using namespace std;

typedef pair<int, int> TPair;

// get the maximum length if bring obj is "st"
int bfs(vector<vector<TPair>> &data, int st)
{
    int max = -1;
    vector<int> visited(data.size(), -1);

    queue<int> q;
    visited[st] = 0;
    q.push(
 
st);
    while (!q.empty())
    {
        auto node = q.front();
        q.pop();
        for (auto r : data[node])
        {
            // important 1. if this flag have been visited before
            if (visited[r.first] != -1)
            {
                // but we still need to judge it. if it is lower when use "node -> r"
 

                // then we still need to push it!!!
                if ((visited[node] + r.second) < visited[r.first])
                {
                    visited[r.first] = visited[node] + r.second;
                    q.push(r.first);
                }
            }
            // important 2. have visited it before or not.
            else
            {
                visited[r.first] = visited[node] + r.second;
                q.push(r.first);
            }
        }
    }
    for (auto r : visited)
    {
        // if someone is -1, it means that can not get this obj!, so print 0 as the question say in the last.
        if (r == -1)
            return -1;
        else
            max = std::max(max, r);
    }
    // debug code
    // cout << st << " " << max << endl;
    return max;
}

int main()
{
    // 1->n transform to 0->n-1
    int n, e;
    cin >> n >> e;
    vector<vector<TPair>> data;
    data.resize(n);

    if (n == 0)
        cout << 0 << endl;

    // load the raw data, warning: flag transformation!
    for (auto i = 0; i < e; i++)
    {
        int a, b, c;
        cin >> a >> b >> c;
        data[a - 1].push_back({b - 1, c});
        data[b - 1].push_back({a - 1, c});
    }

    auto min = bfs(data, 0);
    auto minFlag = 0;
    for (auto i = 1; i < n; i++)
    {
        auto res = bfs(data, i);
        if (res == -1)
        {
            cout << 0 << endl;
            return 0;
        }

        if (res < min)
        {
            minFlag = i;
            min = res;
        }
    }

    cout << minFlag + 1 << " " << min << endl;

    return 0;
}