POJ2288 Islands and bridges 【状态压缩，计数】

分成两问。第一问求Hamilton路径最大赋值，第二问求值最大的hamilton路径有多少个。使用状态压缩来记录状态简化转移。

把细节关注好是重点。本题代码比较繁琐，一定要养成良好的代码习惯，避免出现奇怪的bug。

第一份代码，用bfs实现记忆化搜索。

#include <queue>
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

#define int long long

const int N = 14;
const int INF = 0x3f3f3f3f3f3f3f3f;


int T, n, m, val[N], res[N][N][1 << N], cnt[N][N][1 << N];

bool G[N][N];

void Init () {
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            G[i][j] = 0;
            for (int k = 0; k < (1 << N); ++k) {
                res[i][j][k] = -INF;
                cnt[i][j][k] = 0;
            }
        }
    }
}

struct Node {
    int cur, pre, sit;
};

queue <Node> que;

signed main () {
    // freopen ("data.in", "r", stdin);
    cin >> T;
    while (T--) {
        Init ();
        cin >> n >> m;
        for (int i = 1; i <= n; ++i) {
            cin >> val[i];
        }
        for (int i = 1; i <= m; ++i) {
            int u, v;
            cin >> u >> v;
            G[u][v] = G[v][u] = true;
        }
        if (n == 1) {
            cout << val[1] << " 1" << endl;
            continue;
        }
        for (int i = 1; i <= n; ++i) {
            res[i][0][1 << (i - 1)] = 0;
            cnt[i][0][1 << (i - 1)] = 1;
            que.push ((Node) {i, 0, 1 << (i - 1)}); 
        }
        while (!que.empty ()) {
            Node _node = que.front ();
            que.pop ();
            // printf ("Node now: {cur = %d, pre = %d, sit = %d, res = %d}
", _node.cur, _node.pre, _node.sit, res[_node.cur][_node.pre][_node.sit]);
            for (int i = 1; i <= n; ++i) {
                if (G[_node.cur][i] == false) {
                    continue;
                }
                if ((_node.sit & (1 << (i - 1))) == 0) {
                    // printf ("%d : yes!
", i);
                    // 第二位尚未到达
                    int _newres = res[_node.cur][_node.pre][_node.sit] + val[_node.cur] * val[i];
                    if (G[i][_node.pre]) {
                        _newres += val[_node.pre] * val[_node.cur] * val[i];
                    } 
                    int _newcur = i;
                    int _newpre = _node.cur;
                    int _newsit = _node.sit | (1 << (i - 1));
                    // printf ("Node next: {cur = %d, pre = %d, sit = %d, res = %d}
", _newcur, _newpre, _newsit, _newres);
                    if (_newres > res[_newcur][_newpre][_newsit]) {
                        cnt[_newcur][_newpre][_newsit] = cnt[_node.cur][_node.pre][_node.sit];
                        // printf ("cnt next = %d
", cnt[_newcur][_newpre][_newsit]);
                        if (res[_newcur][_newpre][_newsit] < 0) {
                            que.push ((Node) {_newcur, _newpre, _newsit});
                        }
                        res[_newcur][_newpre][_newsit] = _newres;
                    } else if (_newres == res[_newcur][_newpre][_newsit]) {
                        // printf ("cnt next += %d
", cnt[_node.cur][_node.pre][_node.sit]);
                        cnt[_newcur][_newpre][_newsit] += cnt[_node.cur][_node.pre][_node.sit];
                    }
                }
            }
        }
        int ans1 = -INF, ans2 = 0, sum = 0;
        for (int i = 1; i <= n; ++i) {
            sum += val[i];
            for (int j = 1; j <= n; ++j) {
                if (ans1 < res[i][j][(1 << n) - 1]) {
                    ans1 = res[i][j][(1 << n) - 1];
                    ans2 = cnt[i][j][(1 << n) - 1];
                } else if (ans1 == res[i][j][(1 << n) - 1]) {
                    ans2 += cnt[i][j][(1 << n) - 1];
                }
                // cout << ans2 << endl;
            }
        }
        if (ans1 < 0) {
            cout << "0 0" << endl;
        } else {
            cout << ans1 + sum << " " << ans2 / 2 << endl;
        }
    }
}

第二份代码，使用循环的形式。

#include <queue>
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;

#define int long long

const int N = 14;
const int INF = 0x3f3f3f3f3f3f3f3f;


int T, n, m, val[N], res[N][N][1 << N], cnt[N][N][1 << N];

bool G[N][N];

void Init () {
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            G[i][j] = false;
            for (int k = 0; k < (1 << N); ++k) {
                res[i][j][k] = -INF;
                cnt[i][j][k] = 0;
            }
        }
    }
}

signed main () {
    cin >> T;
    while (T--) {
        Init ();
        cin >> n >> m;
        for (int i = 1; i <= n; ++i) {
            cin >> val[i];
        }
        for (int i = 1; i <= m; ++i) {
            int u, v;
            cin >> u >> v;
            G[u][v] = G[v][u] = true;
        }
        if (n == 1) {
            cout << val[1] << " 1" << endl;
            continue;
        }
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (i == j || !G[i][j]) {
                    continue;
                }
                int s1 = 1 << (i - 1);
                int s2 = 1 << (j - 1);
                res[i][j][s1 ^ s2] = val[i] * val[j];
                cnt[i][j][s1 ^ s2] = 1;
            }
        }
        for (int sit = 0; sit < 1 << n; ++sit) {
            for (int u = 1; u <= n; ++u) {
                int s1 = 1 << (u - 1);
                if (!sit & s1) continue;
                for (int v = 1; v <= n; ++v) {
                    int s2 = 1 << (v - 1);
                    if (u == v || (!G[u][v]) || (!(sit & s2)) || res[v][u][sit] == -INF) continue;
                    for (int t = 1; t <= n; ++t) {
                        int s3 = 1 << (t - 1);
                        if (t == u || t == v || (sit & s3) || !G[v][t]) continue;
                        int sum = val[v] * val[t];
                        if (G[u][t]) {
                            sum += val[u] * val[v] * val[t];
                        }
                        if (res[v][u][sit] + sum > res[t][v][sit ^ s3]) {
                            res[t][v][sit ^ s3] = res[v][u][sit] + sum;
                            cnt[t][v][sit ^ s3] = cnt[v][u][sit];
                        } else if (res[v][u][sit] + sum == res[t][v][sit ^ s3]) {
                            cnt[t][v][sit ^ s3] += cnt[v][u][sit];
                        }
                    }
                }
            }
        }
        int ans1 = -INF, ans2 = 0, sum = 0;;
        for (int i = 1; i <= n; ++i) {
            sum += val[i];
            for (int j = 1; j <= n; ++j) {
                if (ans1 < res[i][j][(1 << n) - 1]) {
                    ans1 = res[i][j][(1 << n) - 1];
                    ans2 = cnt[i][j][(1 << n) - 1];
                } else if (ans1 == res[i][j][(1 << n) - 1]) {
                    ans2 += cnt[i][j][(1 << n) - 1];
                }
            }
        }
        if (ans1 < 0) {
            cout << "0 0" << endl;
        } else {
            cout << ans1 + sum << " " << ans2 / 2 << endl;
        }
    }
}