状态压缩DP题单

P1433 吃奶酪（最短路）

dp(i, s) 表示从 i 出发经过的点的记录为 s 的路线距离最小值

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N = 20;
signed main()
{	
	int n; cin >> n;
	vector<double>x(n + 1), y(n + 1);
	vector<vector<double>> dp(n + 1, vector<double>(1 << (n + 1), 2e9));
	x[0] = 0, y[0] = 0;

	auto dis = [&](int i, int j) -> double
	{
		return sqrt((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]));
	};
	for(int i = 1; i <= n; i ++)
	{
		cin >> x[i] >> y[i];
//		dp[i][1 << (i - 1)] = 0;
	}
	dp[0][1] = 0;
	for(int s = 1; s < (1 << (n + 1)); s ++)
	{
		for(int i = 0; i <= n; i ++)
		{
			if((s & (1 << (i))) == 0) continue;
			for(int j = 0; j <= n; j ++)
			{
				if(i == j) continue;
				if((s & (1 << (j))) == 0) {continue;}
//				cout << s << " " << i << " " << j << '\n';
				dp[i][s] = min(dp[i][s], dp[j][s - (1 << (i))] + dis(i, j)); 
			}
		}
	}
    double ans = -1;
    for(int i = 1; i <= n; i ++)
    {
        double s = dp[i][(1 << (n + 1)) -1];
        if(ans==-1||ans>s) ans=s;
    }
	cout << fixed << setprecision(2) << ans << '\n';
}

P8733 [蓝桥杯 2020 国 C] 补给（floyd）

dp(i, s) 表示到达 i ，经过的点的记录为 s 的路线距离最小值

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N = 20;
signed main()
{	
	int n, D; cin >> n >> D;
	vector<double>x(n + 1), y(n + 1);
	vector<vector<double>> dp(n + 1, vector<double>((1 << n) + 10, 2e9)), f(n + 1, vector<double>(n + 1, 2e9));
	auto dis = [&](int i, int j) -> double
	{
		return sqrt((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]));
	};
	for(int i = 0; i < n; i ++) cin >> x[i] >> y[i];
	
	for(int i = 0; i < n; i ++)
		for(int j = 0; j < n; j ++)
		{
			double now = dis(i, j);
			if(now <= D) f[i][j] = now;
		}
	
	for(int k = 0; k < n; k ++)
		for(int i = 0; i < n; i ++)
			for(int j = 0; j < n; j ++)
				f[i][j] = min(f[i][k] + f[k][j], f[i][j]);

	dp[0][1] = 0;
	for(int s = 1; s < (1 << n); s ++)
	{
		for(int i = 0; i < n; i ++)
		{
			if((s & (1 << i)) == 0) continue;
			for(int j = 0; j < n; j ++)
			{	
				if(i == j) continue;
				if((s & (1 << j)) == 0) continue;
				dp[i][s] = min(dp[i][s], dp[j][s - (1 << i)] + f[i][j]);
//				cout << i << " " << j << " " << f[i][j] << '\n';
			}
		}
	}
	double ans = 1e9;
	for(int i = 0; i < n; i ++)
	{
//		cout << i << " " << (1 << (n - 1)) << " " << dp[i][1 << (n - 1)] << '\n';
		double now = f[0][i] + dp[i][(1 << n) - 1];
		ans = min(ans, now);
	}
	cout << fixed << setprecision(2) << ans << '\n';
}

P7859 [COCI2015-2016#2] GEPPETTO（简单枚举状态）

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N = 4e5 + 10;
typedef long long ll;
int a[25][25]; 
signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int n, m; cin >> n >> m;
	for(int i = 1; i <= m; i ++)
	{
		int x, y; cin >> x >> y;
		a[x][y] = a[y][x] = 1;
	}
	int ans = 1;
	for(int s = 1; s < (1 << n); s ++)
	{
		int f = 0;
		for(int i = 1; i <= n; i ++)
		{
			if((s & (1 << (i - 1))) == 0) continue; 
			for(int j = i + 1; j <= n; j ++)
			{
				if((s & (1 << (j - 1))) == 0) continue;
				if(a[i][j] || a[j][i]) 
				{
					f = 1; 
//					cout << s << " " << i << " " << (s & (1 << (i - 1))) << " " << j << " "<< (s & (1 << (j - 1))) << '\n'; 
					break;
				}
			}
			if(f) break;
		}
//		cout << s << ' ' << f << '\n';
		if(!f) ans ++;
	}
	cout << ans << '\n';
}

P8687 [蓝桥杯 2019 省 A] 糖果

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N = 20;
/*
1 3 3
1 2 3
*/
signed main()
{	
	int n, m, k; cin >> n >> m >> k;
	vector<int>dp(1 << 21, -1), a(105);
	for(int i = 1; i <= n; i ++) 
	{
		int s = 0;
		for(int j = 1; j <= k; j ++)
		{
			int x; cin >> x;
			x --;
			s |= (1 << x);
		}
//		s |= (1 << x), s |= (1 << y), s |= (1 << z);
//		int s = (1 << x) + (1 << y) + (1 << z);
		a[i] = s;
	}
	dp[0] = 0;
	for(int s = 0; s < (1 << m); s ++)
	{
		if(dp[s] == -1) continue;
		for(int i = 1; i <= n; i ++)
		{
			if(dp[s | a[i]] == -1 || dp[s] + 1 < dp[s | a[i]])
			{
				dp[s | a[i]] = dp[s] + 1;
//				cout << s << " " << i << " " << a[i] << ' ' << dp[s | a[i]] << '\n';
			}
		}
	}
	cout << dp[(1 << m) - 1] << '\n';
}

[ABC332E] Lucky bag（SOS DP）

dp(s, i) 表示状态为 s 装入 i 个背包的最小值。

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N = 25; 
signed main()
{	
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	int n, d; cin >> n >> d;
	vector<double>a(n + 1);
	vector<vector<double>>dp(1 << n, vector<double>(d + 1, 1e18));
	double sum = 0;
	for(int i = 0; i < n; i ++)
	{
		cin >> a[i];
//		cout << a[i] << ' ';
		sum += a[i];
	}
	double ave = sum / d;
//	cout << ave << '\n';
	for(int s = 0; s < (1 << n); s ++)
	{
		double res = 0;
		for(int j = 0; j < n; j ++) if((1 << j) & s) res += a[j];
		dp[s][1] = (res - ave) * (res - ave);
	}
	for(int i = 2; i <= d; i ++)
	{
		for(int s = 0; s < (1 << n); s ++)
		{
			for(int t = s; t > 0; t = (t - 1) & s)
			{
				dp[s][i] = min(dp[s][i], dp[t][i - 1] + dp[s ^ t][1]);
//				cout << s << " " << i << '\n';
			}
		}		
	}
	cout << fixed << setprecision(15) << dp[(1 << n) - 1][d] / d << '\n';
}