Spring outing in the suburbs

Suburban spring outing

Meaning:
give $n$ places,$m$ roads, each road has a weight$v$， R R Where$R$ want to go, ask how to arrange the itinerary to minimize the cost.

Idea:
First we use $Floyd(O(n^3\left)\right)$Find the minimum cost of each two places.
Because ofRRThe value range of$R$ is$[2,15]$ , we can use binary bits$\left(0or1\right)$ to indicate whether a location has been visited, for example,$6=(0110{)}_2$, It means that we have visited the first and second places.
Then we use $dp\left[i\right]\left[j\right]$ to represent$i,$The minimum cost in$j$ state. Where$i$ is the binary representation mentioned above,$j$ means at the moment$j$ positions (note dp initialization).
Now is the state transition:
$$dp[i+(1<<k)][k]=min(dp[i+1<<k][k],dp[i][j]+F[g\left[j\right]\left[g\right[k\left]\right]$$
The meaning of the equation is whether to pass the state$dp[i][j]$达到$dp[i+(1<<k\left)\right]\left[k\right]$ state (choose the less expensive one). Where$i+(1<<k)$ , i does not pass the kth place, when$i+(1<<k)$ after the first$k$ positions will become$1$。

We will $i\; is$ in the interval$[1,2^r-1]$ Enumerate, enumerate each selection state. $j$ and$k$ is in the interval$[0,r]$上枚举。
最后取$max(dp[(1<<r)-1][i])(i\in [0,r])$

$Code$

#include<bits/stdc++.h>
using namespace std;
const int inf = 1<<23;
const int N = 20;
const int M = 210;
int g[N], F[M][M], dp[1 << 15][20];
int main() {
    
    
    int n, m, r, a, b, c;
    cin >> n >> m >> r;
    for(int i=0; i<r; i++) cin >> g[i];
    //初始化F数组
    for(int i=1; i<=n; i++) {
    
    
        for(int j=1; j<=n; j++) {
    
    
            if(i != j) F[i][j] = inf;
        }
    }
    //读入每条路径
    for(int i=0; i<m; i++) {
    
    
        cin >> a >> b >> c;
        F[a][b] = min(F[a][b], c);
        F[b][a] = F[a][b];
    }
    //floyd求两点之间最小花费
    for(int i=1; i<=n; i++) {
    
    
        for(int j=1; j<=n; j++) {
    
    
            for(int k=1; k<=n; k++) {
    
    
                F[j][k] = min(F[j][k], F[j][i] + F[i][k]);
            }
        }
    }
    //初始化dp数组
    memset(dp, 0x7f, sizeof dp);
    for(int i=0; i<r; i++) {
    
    
        dp[1 << i][i] = 0;
    }
    //枚举i，j，k。状态转移
    for(int i=1; i<(1 << r)-1; i++) {
    
    
        for(int j=0; j<r; j++) {
    
    
            if(i & (1 << j)) {
    
    
                for(int k=0; k<r; k++) {
    
    
                    if(!(i & (1 << k))) 
                    dp[i + (1 << k)][k] = min(dp[i + (1 << k)][k], dp[i][j] + F[g[j]][g[k]]);
                }
            }
        }
    }
    //取最大值，得到答案。
    int ans = inf;
    for(int i=0; i<r; i++) {
    
    
        ans = min(ans, dp[(1 << r)-1][i]);
    }
    cout << ans;
    return 0;
}