The meaning of problems
Face problems too long, self-transfer ...
analysis
At first glance pruning is a search, but is solution-like positive pressure DP
We set may reach a point in binary, set $ d [i] [s] $ represents the shortest path to the point $ I $ length from point set $ s $, $ f [i] [s] $ denotes Add $ I $ depth point after point set $ s $ required minimum cost (depth points to the starting point after the minimum)
First, an enumeration process $ s $ $ $ D array, and then is sequentially updated in accordance with the depth of $ f $ array, the depth of each enumeration $ s $, whichever directly to a subset of all of its minimum cost, the final answer
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <cmath> using namespace std; #define ll long long #define inf 6000000 #define N 13 int n, m, u, v, w, all; int g[N][N], f[N][1 << N], d[N][1 << N]; int main() { scanf("%d%d", &n, &m); all = (1 << n) - 1; for (int i = 1; i <= n; i++) for (int j = 0; j <= all; j++) d[i][j] = f[i][j] = inf; for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) g[i][j] = inf; for (int i = 1; i <= n; i++) g[i][i] = 0; for (int i = 1; i <= m; i++) { scanf("%d%d%d", &u, &v, &w); g[u][v] = g[v][u] = min(g[u][v], w); } for (int s = 0; s <= all; s++) for (int i = 1; i <= n; i++) if ((1 << (i - 1)) & s) for (int j = 1; j <= n; j++) if (!((1 << (j - 1)) & s)) d[j][s] = min(d[j][s], g[i][j]); for (int i = 1; i <= n; i++) f[1][1 << (i - 1)] = 0; for (int i = 2; i <= n; i++) for (int s = 0; s <= all; s++) for (int k = s; k; k = (k - 1) & s) { int tot = 0; for (int j = 1; j <= n; j++) if ((1 << (j - 1)) & (k ^ s)) tot += d[j][k]; f[i][s] = min(f[i][s], f[i - 1][k] + (i - 1) * tot); } printf("%d\n", f[n][all]); return 0; }