Solution Notes: We first use Dijkstra's algorithm to compute the shortest path from each market to each town. Then for each prospective town x in which Farmer John might build his house, we check all possible K! permutations of the markets that could end up being a feasible daily schedule (checking each one is fast since we have now computed all the relevant market-town shortest path distances). Among all these, we remember the best solution.

``````
#include <iostream>
#include <fstream>
#include <queue>
#include <vector>
#include <map>
#include <cstdlib>
#include <cmath>
#include <functional>
#include <cstring>
#include <algorithm>

#define pii pair<int,int>

using namespace std;

int N,M,K;
int markets;
int inf = 1 << 29;
vector<pii> graph; //L then end
int shortest; //shortest path from the ith market to the jth town
bool isMarket; //is town i a market?

void dijkstra (int start) //from a market
{
priority_queue <pii, vector<pii>, greater<pii> > pq;
pq.push(pii(0, markets[start]));

while(!pq.empty()) //standard heap dijkstra
{
int curdist = pq.top().first;
int curnode = pq.top().second;
pq.pop();

if(shortest[start][curnode] <= curdist)
continue;

shortest[start][curnode] = curdist;

for (int i = 0; i < graph[curnode].size(); i++)
{
int nextnode = graph[curnode][i].second;
int nextdist = graph[curnode][i].first + curdist;

if(nextdist < shortest[start][nextnode])
pq.push(pii(nextdist, nextnode));
}
}
}

int main()
{
ifstream in ("relocate.in");
ofstream out ("relocate.out");

in >> N >> M >> K;
for (int i = 0; i < N; i++)
isMarket[i] = false;
for (int i = 0; i < K; i++)
{
in >> markets[i];
markets[i]--;
isMarket[markets[i]] = true;
}
for (int i = 0; i < M; i++)
{
int a,b,L;
in >> a >> b >> L;
a--; b--;
graph[a].push_back(pii(L, b));
graph[b].push_back(pii(L, a));
}

for (int i = 0; i < K; i++)
{
for (int j = 0; j < N; j++)
shortest[i][j] = inf;
dijkstra(i);
}

int best = inf;
int order[K];
for (int i = 0; i < K; i++)
order[i] = i;

//loop over all permutations in which the K markets are visited
do{
int total = inf;
for (int i = 0; i < N; i++) //choose the farm location to minimize the sum of the distances from the farm to the first market and the last market to the farm
if(!isMarket[i])
total = min(total, shortest[order][i] + shortest[order[K-1]][i]);

for (int i = 1; i < K; i++) //add up distances between pairs of markets
total += shortest[order[i-1]][markets[order[i]]];

best = min(best, total);
}while(next_permutation(order, order + K));

out << best << "\n";
out.close();
return 0;
}

``````