一类哈密顿路径/回路为背景的状压dp

https://codeforces.com/contest/1950/problem/G

在非连通图上找到一条包含点最多的路径，dp数组维护可达性

// Problem: G. Shuffling Songs
// Contest: Codeforces - Codeforces Round 937 (Div. 4)
// URL: https://codeforces.com/contest/1950/problem/G
// Memory Limit: 256 MB
// Time Limit: 3000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
using namespace std;
#define ll long long
//# define int long long
#define ull unsigned long long
#define pii pair<int,int>

#define baoliu(x, y) cout << fixed << setprecision(y) << x
#define endl  "\n"
#define debug1(x) cerr<<x<<" "
#define  debug2(x) cerr<<x<<endl
const int N = 2e5 + 10;
const int M = 1e6 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-8;
int n, m;
int a[N];

void solve(){
	vector<string>v1;
	vector<string>v2;
		cin>>n;
	vector<vector<int>>w(n+1,vector<int>(n+1,0));
	
	for(int i=0;i<n;i++){
		string s1,s2;cin>>s1>>s2;
		v1.push_back(s1);
		v2.push_back(s2);
	}
	for(int i=0;i<n;i++){
		for(int j=i;j<n;j++){
			//if(j==i)continue;
			if(v1[i]==v1[j]||v2[i]==v2[j]){
				w[i][j]=1;w[j][i]=1;
			}
		}
	}
	// for(int i=0;i<n;i++){
		// for(int j=0;j<n;j++){
			// cerr<<w[i][j];
		// }
		// cerr<<endl;
	// }
	//寻找答案状态：枚举最终的状态和终点，从而计算答案
	int ans=0;
	vector<vector<int>>dp((1<<n)+1,vector<int>(n+1,0));
	//fill(dp[0].begin(),dp[0].end(),1);
	for(int i=0;i<n;i++)dp[1<<i][i]=1;
	for(int i=0;i<(1<<n);i++){
		for(int j=0;j<n;j++){
			int u=(i>>j)&1;
			if(u==0)continue;
			for(int k=0;k<n;k++){
				//为什么不能判断终点j和k转移重复
				//这会导致初始只有一个0基础态无法转移
				int v=(i>>k)&1;
				if(v==0)continue;
				int last=i-(1<<j);
	dp[i][j]|=dp[last][k]&w[k][j];
//	cerr<<i<<" "<<j<<endl;
	if(dp[i][j]){
		ans=max(ans,__builtin_popcount(i));
	}
	
			}
		}
	}
	
	cout<<n-ans<<endl;
}
int main() {
    cin.tie(0);
    
    ios::sync_with_stdio(false);

    int t;
   cin>>t;
     //t=1;
    while (t--) {
solve();
    }
    return 0;
}

最短哈密顿路径

https://www.acwing.com/problem/content/93/

主要就是需要不重不漏的走一遍并且维护最短路，这无法在多项式时间内完成，在n不大的时候我们用二级制枚举所有状态，枚举终点作为状态，转移是通过枚举上一个点。

时间复杂度：O(\(2^n*n^2\))

// Problem: 最短Hamilton路径
// Contest: AcWing
// URL: https://www.acwing.com/problem/content/93/
// Memory Limit: 256 MB
// Time Limit: 5000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
using namespace std;
#define ll long long
//# define int long long
#define ull unsigned long long
#define pii pair<int,int>

#define baoliu(x, y) cout << fixed << setprecision(y) << x
#define endl  "\n"
#define debug1(x) cerr<<x<<" "
#define  debug2(x) cerr<<x<<endl
const int N = 21;
const int M = 1e6 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-8;
int n, m;
int w[N][N];
//首先思考为什么这里最短路不能用普通的最短路
//首先它需要把所有点都遍历到，而普通最短路根本不在乎这个
int dp[1<<20][N];
void solve(){
	cin>>n;
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			cin>>w[i][j];
		}
	}
	
	memset(dp,0x3f ,sizeof dp);
	dp[1][0]=0;
	for(int i=0;i<(1<<n);i++){
		for(int j=0;j<n;j++){//枚举当前终点
			if(((i>>j)&1)==0)continue;
			for(int k=0;k<n;k++){//枚举上一次转移点
				if(((i>>k)&1)==0)continue;
				int pre=i-(1<<j);
				if(k==j)continue;
				dp[i][j]=min(dp[i][j],dp[pre][k]+w[k][j]);
			}
		}
	}
	int ed=(1<<n)-1;
	cout<<dp[ed][n-1];
}
int main() {
    cin.tie(0);
    
    ios::sync_with_stdio(false);

    int t;
   //cin>>t;
     t=1;
    while (t--) {
solve();
    }
    return 0;
}

最短哈密顿回路

https://www.luogu.com.cn/problem/P1171

// Problem: P1171 售货员的难题
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P1171
// Memory Limit: 125 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
using namespace std;
#define ll long long
//# define int long long
#define ull unsigned long long
#define pii pair<int,int>

#define baoliu(x, y) cout << fixed << setprecision(y) << x
#define endl  "\n"
#define debug1(x) cerr<<x<<" "
#define  debug2(x) cerr<<x<<endl
const int N =21 ;
const int M = 1e6 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-8;
int n, m;
int a[N];

int w[N][N];
//首先思考为什么这里最短路不能用普通的最短路
//首先它需要把所有点都遍历到，而普通最短路根本不在乎这个
int dp[1<<20][N];
void solve(){
    cin>>n;
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            cin>>w[i][j];
        }
    }

    memset(dp,0x3f ,sizeof dp);
    dp[1][0]=0;
    for(int i=0;i<(1<<n);i++){
        for(int j=0;j<n;j++){//枚举当前终点
            if(((i>>j)&1)==0)continue;
            for(int k=0;k<n;k++){//枚举上一次转移点
                if(((i>>k)&1)==0)continue;
                int pre=i-(1<<j);
                if(k==j)continue;
                dp[i][j]=min(dp[i][j],dp[pre][k]+w[k][j]);
            }
        }
    }
    int ed=(1<<n)-1;
    int ans=inf;
    for(int i=1;i<n;i++){
    	ans=min(ans,dp[ed][i]+w[i][0]);
    }
    cout<<ans<<endl;
}


int main() {
    cin.tie(0);
    
    ios::sync_with_stdio(false);

    int t;
   //cin>>t;
     t=1;
    while (t--) {
solve();
    }
    return 0;
}