洛谷__P1171 售货员的难题

题目链接:P1171 售货员的难题 - 洛谷

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题目大意：

有 $n$ 个村庄，要从 村庄1 出发到每个村庄各一次再回到 村庄1

求：最短路程

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代码：

普通遍历：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<queue>
#include<deque>
#include<stack>
#include<set>
#include<map>
#include<unordered_set>
#include<unordered_map>
#include<bitset>
#include<tuple>
#define inf 72340172838076673
#define int long long
#define endl '\n'
#define F first
#define S second

#define  mst(a,x) memset(a,x,sizeof (a))
using namespace std;
typedef pair<int, int> pii;

const int N = 21, mod = 998244353;

int n;
int f[1 << N][N];
int e[N][N];

void solve() {

    cin >> n;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            cin >> e[i][j];
        }
    }
    
    for (int i = 0; i <= (1 << n) -1 ; i++)
        for (int j = 0; j <= n; j++)
            f[i][j] = inf;
            
    f[1][1] = 0;
      
    for (int st = 1; st <= (1 << n) -1; st++) {
    
        for (int u = 1; u <= n; u++) {
        
            if ((st & (1 << u - 1)) == 0) continue;//u点必须走过
            
            for (int v = 1; v <= n; v++) {
                if (st & (1 << v - 1)) continue;//v点必须没走过
                //从状态为st的点u 到 v
                f[st | (1 << v - 1)][v] = min(f[st | (1 << v - 1)][v], f[st][u] + e[u][v]);
            }
        }
        
    }
    
    int res = inf;
    for (int i = 2; i <= n; i++) {
        res = min(res, f[(1 << n) -1][i] + e[i][1]);
    }
    cout << res << endl;
    
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    
    int T = 1;
// cin >> T;
    while (T--) solve();
    
    return 0;
}

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快速的遍历方法：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<queue>
#include<deque>
#include<stack>
#include<set>
#include<map>
#include<unordered_set>
#include<unordered_map>
#include<bitset>
#include<tuple>
#define inf 72340172838076673
#define int long long
#define endl '\n'
#define F first
#define S second

#define  mst(a,x) memset(a,x,sizeof (a))
using namespace std;
typedef pair<int, int> pii;

const int N = 21, mod = 998244353;

int n;
int f[(1 << N) +1][N + 1];//状态:访问过的村庄 要去的村庄
int e[N][N];

void solve() {

    cin >> n;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            cin >> e[i][j];
        }
    }
    
    //直接赋值比memset快
    for (int i = 1; i <= (1 << n) -1; i++) {
        for (int j = 1; j <= n; j++) {
            f[i][j] = inf;
        }
    }
    
    f[1][1] = 0;//初始化
    
    for (int st = 1; st <= (1 << n) -1; st++) {//枚举每个状态
    
        for (int i = st; i; i &= i - 1) {
            //↑ 消除最低位 '1'
            int u = __builtin_ctz(i) +1; //最低位的 '1' 的位置
            
            if (f[st][u] > 1e10) continue;
            
            for (int j = ((1 << n) -1) & ~st; j; j &= j - 1) {
                // ↑ 对当前状态取反
                int v = __builtin_ctz(j) +1;
                //从 u 这个位置出发到 v
                f[st | (1 << v - 1)][v] = min(f[st | (1 << v - 1)][v], f[st][u] + e[u][v]);
            }
        }
        
    }
    
    int res = inf;
    for (int i = 2; i <= n; i++) res = min(res, f[(1 << n) -1][i] + e[i][1]);
    cout << res << endl;
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    
    int T = 1;
// cin >> T;
    while (T--) solve();
    
    return 0;
}

dfs但TLE#9：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<queue>
#include<deque>
#include<stack>
#include<set>
#include<map>
#include<unordered_set>
#include<unordered_map>
#include<bitset>
#include<tuple>
#define inf 72340172838076673
#define int long long
#define endl '\n'
#define F first
#define S second
#define  mst(a,x) memset(a,x,sizeof (a))
using namespace std;
typedef pair<int, int> pii;

const int N = 21, mod = 998244353;

int n;
int f[N][1 << N];
int e[N][N];

void dfs(int u, int path) {

    for (int i = 1; i <= n; i++) {
    
        int now = path | 1 << i - 1 ;
        if ((1 << i - 1) & path) continue;
        
        if (f[i][now] > f[u][path] + e[u][i]) {
            f[i][now] = f[u][path] + e[u][i];
            dfs(i, now);
        }

    }
    
}

void solve() {

    cin >> n;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            cin >> e[i][j];
        }
    }
    
    mst(f, 1);
    f[1][1] = 0;
    dfs(1, 1);
    
    int res = inf;
    for (int i = 1; i <= n ; i++) {
        res = min(res, f[i][(1 << n) -1] + e[i][1]);
    }
    
    cout << res << endl;
    
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    
    int T = 1;
// cin >> T;
    while (T--) solve();
    
    return 0;
}