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

const int N = 800, M = 3000, INF = 0x3f3f3f3f;

int h[N], e[M], w[M], ne[M], idx;
int id[N];
bool st[N];
int dist[N];
int n, p, c;

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

int spfa(int start)
{
    memset(dist, 0x3f, sizeof dist);
    queue<int> q;
    q.push(start);
    st[start] = true;
    dist[start] = 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;
                }
            }
        }
    }

    int res = 0;
    for (int i = 1; i <= n; i ++)
    {
        int j = id[i];
        if (dist[j] == INF) return INF;
        res += dist[j];
    }
    return res;
}

int main()
{

    cin >> n >> p >> c;

    memset(h, -1, sizeof h);
    for (int i = 1; i <= n; i ++) cin >> id[i];
    for (int i = 1; i <= c; i ++)
    {
        int a, b, w;
        cin >> a >> b >> w;
        add(a, b, w);
        add(b, a, w);
    }

    int res = INF;

    for (int i = 1; i <= p; i ++)
    {
        res = min(res, spfa(i));
    }
    cout << res;
    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