Problem:
Score : 900 points

Problem Statement

Mole decided to live in an abandoned mine. The structure of the mine is represented by a simple connected undirected graph which consists of N vertices numbered 1 through N and M edges.
The i-th edge connects Vertices a_i and b_i, and it costs c_i yen (the currency of Japan) to remove it.

Mole would like to remove some of the edges so that there is exactly one path from Vertex 1 to Vertex N that does not visit the same vertex more than once. Find the minimum budget needed to achieve this.

Constraints

2 \leq N \leq 15

N-1 \leq M \leq N(N-1)/2

1 \leq a_i, b_i \leq N

1 \leq c_i \leq 10^{6}

There are neither multiple edges nor self-loops in the given graph.

The given graph is connected.

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1 c_1
:
a_M b_M c_M

Output

Print the answer.

Sample Input 1

4 6
1 2 100
3 1 100
2 4 100
4 3 100
1 4 100
3 2 100

Sample Output 1

200

By removing the two edges represented by the red dotted lines in the figure below, the objective can be achieved for a cost of 200 yen.

Sample Input 2

2 1
1 2 1

Sample Output 2

0

It is possible that there is already only one path from Vertex 1 to Vertex N in the beginning.

Sample Input 3

15 22
8 13 33418
14 15 55849
7 10 15207
4 6 64328
6 9 86902
15 7 46978
8 14 53526
1 2 8720
14 12 37748
8 3 61543
6 5 32425
4 11 20932
3 12 55123
8 2 45333
9 12 77796
3 9 71922
12 15 70793
2 4 25485
11 6 1436
2 7 81563
7 11 97843
3 1 40491

Sample Output 3

133677

Solution in C++14 (GCC 5.4.1) (judge verdict: Memory Limit Exceeded)


#include <bits/stdc++.h>

#define ln                '\n'
#define all(dat)           dat.begin(), dat.end()
#define loop(i, to)        for (int i = 0; i < to; ++i)
#define cont(i, to)        for (int i = 1; i <= to; ++i)
#define circ(i, fm, to)    for (int i = fm; i <= to; ++i)
#define foreach(i, dat)    for (__typeof(dat.begin()) i = dat.begin(); i != dat.end(); ++i)

typedef long long          num;

using namespace std;

const int nsz = 15, msksz = (1 << nsz) - 1, inf = 0x3f3f3f3f;
int n, m, to, sum, id[msksz + 5], g[nsz + 5][nsz + 5];
int dp[msksz + 5][nsz + 5], ts[msksz + 5][nsz + 5], s[msksz + 5], ans;
map<pair<int, int>, int> mx;

void inline link(int u, int v, int w) {
    g[u][v] = g[v][u] = w;
}

string inline str(int S) {
    string s;
    loop (i, n) {
        s += ((S >> i) & 1) + 48;
    }
    return s;
}

void inline init() {
    loop (i, n) {
        id[1 << i] = i;
    }
}

void inline upd(int &a, int b) {
    a < b && (a = b);
}

int inline qry(int S1, int S2) {
    return mx[make_pair(S1, S2)];
}

int main() {
    ios::sync_with_stdio(0);
    cin >> n >> m;
    init();
    to = (1 << n) - 1;
    cont (i, m) {
        int u, v, w;
        cin >> u >> v >> w;
        --u, --v;
        link(u, v, w);
        sum += w;
    }
    circ (S, 0, to) {
        int sum = 0;
        for (int t = S, d1 = t & -t; d1; t -= d1, d1 = t & -t) {
            int u = id[d1];
            for (int t = S, d2 = t & -t; d2; t -= d2, d2 = t & -t) {
                int v = id[d2];
                sum += (u < v) * g[u][v];
            }
        }
        s[S] = sum;
    }
    circ (S, 0, to) {
        int rS = to ^ S;
        for (int t = rS, d2 = t & -t; d2; t -= d2, d2 = t & -t) {
            int u = id[d2], sum = 0;
            for (int t = S, d1 = t & -t; d1; t -= d1, d1 = t & -t) {
                int v = id[d1];
                sum += g[u][v];
            }
            ts[S][u] = sum;
            // cerr << str(S) << ' ' << u << ": " << sum << ln;
        }
    }
    circ (S, 0, to) {
        int rS = to ^ S;
        for (int tS = rS; tS; tS = rS & (tS - 1)) {
            // if (S >= tS)  continue;
            int w = 0;
            for (int t = S, d1 = t & -t; d1; t -= d1, d1 = t & -t) {
                int u = id[d1];
                upd(w, ts[tS][u]);
                // for (int t = tS, d2 = t & -t; d2; t -= d2, d2 = t & -t) {
                //     int v = id[d2];
                //     // cerr << u << ' ' << v << ln;
                //     upd(w, g[u][v]);
                // }
            }
            mx[make_pair(S, tS)] = w;
            // cerr << str(S) << ' ' << str(tS) << ": " << w << ln;
        }
    }
    // cerr << mx.size() << ln;
    memset(dp, -inf, sizeof(dp));
    dp[1][0] = 0;
    circ (S, 0, to) {
        if (!(S & 1))  continue;
        // cerr << str(S) << ln;
        int rS = to ^ S;
        for (int tS = rS; tS; tS = rS & (tS - 1)) {
            if ((tS >> (n - 1)) & 1)  continue;
            int mx = qry(S, tS);
            // cerr << str(S) << ' ' << str(tS) << ": " << mx << ln;
            // if (mx == 0) {
            //     cerr << str(S) << ' ' << str(tS) << ln;
            // }
            for (int t = S, d = t & -t; d; t -= d, d = t & -t) {
                int u = id[d];
                // cerr << str(S) << ' ' << u << " => " << str(S | tS) << ' ' << u << ' ' << dp[S][u] + mx + s[tS] << ln;
                upd(dp[S | tS][u], dp[S][u] + mx + s[tS]);
            }
        }
        for (int t = S, d1 = t & -t; d1; t -= d1, d1 = t & -t) {
            int u = id[d1];
            for (int t = rS, d2 = t & -t; d2; t -= d2, d2 = t & -t) {
                int v = id[d2];
                if (!g[u][v])  continue;
                if (v == 0)  continue;
                // cerr << d1 << ',' << d2 << ": " << u << ',' << v << ln;
                // cerr << str(S) << ' ' << u << " -> " << str(S | (1 << v)) << ' ' << v << ": " << dp[S][u] + g[u][v] << ln;
                upd(dp[S | (1 << v)][v], dp[S][u] + g[u][v]);
            }
        }
    }
    // circ (S, 0, to) {
    //     if (!(S & 1))  continue;
    //     for (int t = S, d = t & -t; d; t -= d, d = t & -t) {
    //         int u = id[d];
    //         cerr << str(S) << ' ' << u << ": " << dp[S][u] << ln;
    //     }
    // }
    circ (S, 0, to) {
        upd(ans, dp[S][n - 1]);
    }
    cerr << sum << ' ' << ans << ln;
    cout << sum - ans << ln;
    cerr << sum - ans << ln;
}