题目描述

blablabla

样例

blablabla

算法1

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

#include<iostream>
#include<cstring>
#include<queue>
using namespace std;

typedef pair<int, int> PII;

const int N = 50010, M = 200010, INF = 0x3f3f3f3f;

int n, m;
int dist[6][N];
bool st[N];
int relation[6];
int h[N], e[M], ne[M], w[M], idx;

void add(int a, int b, int c)
{
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx ++;
}


void spfa(int source, int dist[])
{
    memset(dist, 0x3f, N * 4);

    queue<int> q;

    q.push(source);
    st[source] = true;
    dist[source] = 0;

    while(q.size())
    {
        auto t = q.front();
        q.pop();
        st[t] = false;

        for (int i = h[t]; ~i; i = ne[i])
        {
            int j = e[i], d = w[i];
            if (dist[j] > dist[t] + d)
            {
                dist[j] = dist[t] + d;
                if (!st[j])
                {
                    q.push(j);
                    st[j] = true;
                }

            }
        }
    }
}

void dijkstra(int source, int dist[])
{
    memset(dist, 0x3f, N * 4);
    memset(st, 0, sizeof st);
    priority_queue<PII, vector<PII>, greater<PII>> heap;

    heap.push({0, source});
    dist[source] = 0;

    while(heap.size())
    {
        auto t = heap.top();
        heap.pop();
        int u = t.second, d = t.first;
        if (st[u]) continue;
        st[u] = true;

        for (int i = h[u]; ~i; i = ne[i])
        {
            int j = e[i];
            if (!st[j] && dist[j] > d + w[i])
            {
                dist[j] = d + w[i];
                heap.push({dist[j], j});
            }
        }
    }
}

int dfs(int u, int last, int distance)
{
    if (u == 6) return distance;
    int res = INF;
    for (int i = 1; i <= 5; i ++ )
    {
      if (!st[i])
      {
          st[i] = true;
          int j = relation[i];
          res = min(res, dfs(u + 1, i, distance + dist[last][j]));
          st[i] = false;
      }
    }
    return res;
}

int main()
{
    cin >> n >> m;
    for (int i = 1; i <= 5; i ++) cin >> relation[i];
    relation[0] = 1;

    memset(h, -1, sizeof h);
    while(m --)
    {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
        add(b, a, c);
    }

   for (int i = 0; i < 6; i ++) dijkstra(relation[i], dist[i]);

    memset(st, 0, sizeof st);
    cout << dfs(1, 0, 0) << endl;
    return 0;
}

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

blablabla