状压DP

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状压DP

状压 DP 是动态规划的一种，通过将状态压缩为整数来达到优化转移的目的。

例题

售货员的难题 洛谷1171

#include<bits/stdc++.h>
using namespace std;
int n,i,j,k,min1,a[25][25],f[1050000][25];
int main()

	cin>>n;
	for (i=1;i<=n;i++)
		for (j=1;j<=n;j++)
			cin>>a[i][j];
	memset(f,64,sizeof(f));
	f[1][1]=0;
	for (i=1;i<1<<n;i+=2)
	
		for (j=1;j<=n;j++)
		
			if (!(i>>(j-1)&1))
				continue;
			for (k=1;k<=n;k++)
			
				if (j==k)
					continue;
				if (!(i>>(k-1)&1)) 
					continue;
				f[i][j]=min(f[i][j],f[i^(1<<(j-1))][k]+a[k][j]);
			
		
	
	min1=2147483647;
	for (i=1;i<=n;i++)
		min1=min(min1,f[(1<<n)-1][i]+a[i][1]);
	cout<<min1<<\'\\n\';

单词游戏 洛谷1278

#include<bits/stdc++.h>
using namespace std;
long long n,i,max1,f[1<<16][5];
string s;
struct ff
	long long bg,ed,len;
a[20];
long long sc(long long x,long long y)

	if (f[x][y]!=-1)
		return f[x][y];
	long long s=0;
	for (int i=0;i<n;i++)
		if ((y==a[i].bg)and(!((1<<i)&x)))
			s=max(s,sc(x|1<<i,a[i].ed)+a[i].len);
	f[x][y]=s;
	return s;

int change(char ch)

	if (ch==\'A\')
		return 0;
	if (ch==\'E\')
		return 1;
	if (ch==\'I\')
		return 2;
	if (ch==\'O\')
		return 3;
	if (ch==\'U\')
		return 4;

int main()

	cin>>n;
	for (i=0;i<n;i++)
	
		cin>>s;
		a[i].bg=change(s[0]);
		a[i].ed=change(s[s.length()-1]);
		a[i].len=s.length();
	
	for (i=0;i<n;i++)
	
		memset(f,-1,sizeof(f));
		max1=max(max1,sc(1<<i,a[i].ed)+a[i].len);
	
	cout<<max1<<\'\\n\';

平板涂色 洛谷1283

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,i,j,k,s,to,min1,f[1<<17][21];bool ff;
struct sc
	ll ax,ay,bx,by,c;
a[40];
vector<ll>b[17];
int main()

	cin>>n;
	for (i=0;i<n;i++)
		cin>>a[i].ax>>a[i].ay>>a[i].bx>>a[i].by>>a[i].c;
	for (i=0;i<n;i++)
		for (j=0;j<n;j++)
			if (i!=j)
			
				if ((a[i].bx==a[j].ax)and((a[i].ay<=a[j].ay)and(a[i].by>=a[j].ay)or(a[i].ay<=a[j].by)and(a[i].by>=a[j].by)or(a[i].ay>=a[j].ay)and(a[i].by<=a[j].by)))
				b[j].push_back(i);
			
	memset(f,64,sizeof(f));
	for (i=0;i<=20;i++)
		f[0][i]=1;
	for (i=0;i<(1<<n);i++)
	
		for (j=0;j<n;j++)
		
			if (!((1<<j)&i))
				continue;
			ff=true;s=1000000;
			for (k=0;k<b[j].size();k++)
			
				to=b[j][k];
				if ((!((1<<to)&i)))
				
					ff=false;
					break;
				
				s=min(s,f[i^(1<<j)][a[to].c]+(!(a[to].c==a[j].c)));
			
			if (ff)
			
				for (k=1;k<=20;k++)
					s=min(s,f[i^(1<<j)][k]+(!(a[j].c==k)));
				f[i][a[j].c]=s;
			
		
	
	min1=1000000;
	for (i=1;i<=20;i++)
		min1=min(min1,f[(1<<n)-1][i]);
	cout<<min1<<\'\\n\';

互不侵犯 洛谷1896

#include<bits/stdc++.h>
using namespace std;
long long n,m,i,j,k,s,q,p,l,f[15][1050][105];
long long judge(long long x)

	int k=0,s=0;
	while (x>0)
	
		if ((x%2==1)and(k==1))
			return -1;
		k=x%2;
		s=s+k;
		x=x/2;
	
	if (s<=m)
		return s;
	else
		return -1;

bool judge2(long long x,long long y)

	int t2=0,w2=0;
	while ((x>0)or(y>0))
	
		int t=x%2;int w=y%2;
		if ((t==1)and((w==1)or(w2==1)))
			return true;
		if ((w==1)and((t==1)or(t2==1)))
			return true;
		t2=t;w2=w;
		x=x/2;y=y/2;
	
	return false;

int main()

	cin>>n>>m;
	for (i=0;i<1<<n;i++)
	
		q=judge(i);
		if ((q!=-1)and(q<=m))
			f[1][i][q]=1;
	
	for (i=1;i<=n;i++)
		for (j=0;j<1<<n;j++)
		
			q=judge(j);
			if (q==-1)	continue;
			for (k=0;k<1<<n;k++)
			
				p=judge(k);
				if ((p==-1)or(q+p>m)or(judge2(j,k))) continue;
				for (l=q;l<=m-p;l++)
					f[i][k][l+p]=f[i][k][l+p]+f[i-1][j][l];
			
		
	for (i=0;i<1<<n;i++)
		s=s+f[n][i][m];
	cout<<s<<\'\\n\';

炮兵阵地 洛谷2704

#include<bits/stdc++.h>
using namespace std;
int n,m,i,j,k,l,ma,a[105],b[1050],f[5][1050][1050];
char ch;
int judge(int x)

	int s=0,k=0,a=0,b=0;
	while (x>0)
	
		k=x%2;s+=k;
		if ((k==1)and((a==1)or(b==1)))
			return -1;
		a=b;
		b=k;
		x=x/2;
	
	return s;

int main()

	cin>>n>>m;
	for (i=1;i<=n;i++)
		for (j=1;j<=m;j++)
		
			cin>>ch;
			if (ch==\'H\')
				a[i]=a[i]*2+1;
			else a[i]=a[i]*2;
		
	for (i=0;i<(1<<m);i++)
	
		b[i]=judge(i);
		if ((!(i&a[1]))and(b[i]!=-1))
			f[1][i][0]=b[i];
	
	for (i=0;i<(1<<m);i++)
		if ((!(i&a[1]))and(b[i]!=-1))
			for (j=0;j<(1<<m);j++)
				if ((!(j&a[2]))and(b[j]!=-1)and(!(i&j)))
					f[2][j][i]=max(f[2][j][i],f[1][i][0]+b[j]);
	for (i=3;i<=n;i++)
		for (j=0;j<(1<<m);j++)
			if ((b[j]!=-1)and(!(j&a[i])))
				for (k=0;k<(1<<m);k++)
					if ((b[k]!=-1) and (!(k&a[i-1])) and (!(j&k)))
						for (l=0;l<(1<<m);l++)
							if ((b[l]!=-1) and (!(l&a[i-2])) and (!(j&l)) and (!(k&l)))
								f[i%3][j][k]=max(f[i%3][j][k],f[(i-1)%3][k][l]+b[j]);
	for (i=0;i<(1<<m);i++)
		for (j=0;j<(1<<m);j++)
			if ((b[i]!=-1)and(b[j]!=-1)and (!(i&j)))
				ma=max(ma,f[n%3][i][j]);
	cout<<ma<<\'\\n\';

Little Pony and Harmony Chest

https://codeforces.com/contest/453/problem/B

#include<bits/stdc++.h>
using namespace std;
int n, m, i, j, k, x, y, t, mi, a[105], b[105], f[105][1 << 18], g[105][1<<18],c[105],d[105];
bool f2;
int main()

	cin >> n;
	for (i = 1; i <= n; i++)
		cin >> a[i];
	for (i = 2; i <= 60; i++)
	
		for (j = 2; j <= i / 2; j++)
			if (i % j == 0)
				break;
		if (j == i / 2 + 1)
			b[++m] = i;
	
	for (i = 1; i <= 60; i++)
		for (j = 1; j <= m; j++)
			if (i % b[j] == 0)
				d[i] = d[i] ^ (1 << (j - 1));
	memset(f, 0x3f, sizeof(f));
	f[0][0] = 0;
	for (i=1;i<=n;i++)
		for (j = 0; j < 1 << m; j++)
			for (k = 1; k <= a[i]*2; k++)
				if ((d[k] & j) == d[k])
				
					t = j ^ d[k];
					if (f[i - 1][t] + abs(a[i] - k) < f[i][j])
					
						f[i][j] = f[i - 1][t] + abs(a[i] - k);
						g[i][j] = k;
					
				
	mi = 1e11;
	for (i = 0; i < 1 << m; i++)
		if (f[n][i] < mi)
		
			mi = f[n][i];
			k = i;
		
	for (i = n; i >= 1; i--)
	
		c[i] = g[i][k];
		x = g[i][k]; y = 1;
		while (x > 1)
		
			f2 = false;
			while (x % b[y] == 0)
			
				x = x / b[y];
				f2 = true;
			
			if (f2)
				k = k ^ (1 << (y - 1));
			y++;
		
	
	for (i = 1; i < n; i++)
		cout << c[i] << \' \';
	cout << c[n] << \'\\n\';

Another Sith Tournament

https://codeforces.com/contest/678/problem/E

#include<bits/stdc++.h>
using namespace std;
int n, i, j, k, l, b[1 << 19];
double a[20][20], f[1 << 19], ma;
int main()

	cin >> n;
	for (i = 1; i <= n; i++)
		for (j = 1; j <= n; j++)
			cin >> a[i][j];
	for (i = 0; i < 1 << n; i++)
	
		k = i; 
		while (k > 0)
		
			b[i] = b[i] + k % 2;
			k = k / 2;
		
	
	f[1] = 1;
	for (i = 2; i <= n; i++)
		for (j = 0; j < 1 << n; j++)
			if (b[j] == i)
				for (l = 1; l <= n; l++)
					if (j & (1 << (l - 1)))
						for (k = 1; k <= n; k++)
							if (j & (1 << (k - 1)))
								if (k != l)
									f[j] = max(f[j], f[j ^ (1 << (l - 1))] * a[k][l] + f[j ^ (1 << (k - 1))] * a[l][k]);
	printf("%.6f", f[(1<<n)-1]);

Binary Table

https://codeforces.com/contest/662/problem/C

#include<bits/stdc++.h>
using namespace std;
int n, m, i, j, k, s, ma; char ch;
int a[22][100005], f[22][1 << 21];
int main()

	cin >> n >> m;
	for (i = 1; i <= n; i++)
		for (j = 1; j <= m; j++)
		
			cin >> ch;
			a[i][j] = ch - \'0\';
		
	for (i = 1; i <= m; i++)
	
		s = 0; 
		for (j = 1; j <= n; j++)
			s = s * 2 + a[j][i];
		f[0][s]++;
	
	for (i = 1; i <= n; i++)
		for (j = n; j >= 1; j--)
			for (k = 0; k < 1 << n; k++)
				f[j][k] = f[j][k] + f[j - 1][k ^ (1 << (i - 1))];
	ma = 1e10;
	for (i = 0; i < (1 << n); i++)
	
		s = 0;
		for (j = 1; j <= n; j++)
			s =	s + f[j][i] * min(j, n - j);
		ma = min(ma, s);
	
	cout << ma << \'\\n\';

fzu2188 状压dp

G - Simple String Problem
Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Description

Recently, you have found your interest in string theory. Here is an interesting question about strings.

You are given a string S of length n consisting of the first k lowercase letters.

You are required to find two non-empty substrings (note that substrings must be consecutive) of S, such that the two substrings don‘t share any same letter. Here comes the question, what is the maximum product of the two substring lengths?

Input

The first line contains an integer T, meaning the number of the cases. 1 <= T <= 50.

For each test case, the first line consists of two integers n and k. (1 <= n <= 2000, 1 <= k <= 16).

The second line is a string of length n, consisting only the first k lowercase letters in the alphabet. For example, when k = 3, it consists of a, b, and c.

Output

For each test case, output the answer of the question.

Sample Input

4
25 5
abcdeabcdeabcdeabcdeabcde
25 5
aaaaabbbbbcccccdddddeeeee
25 5
adcbadcbedbadedcbacbcadbc
3 2
aaa

Sample Output

6 150 21 0

Hint

One possible option for the two chosen substrings for the first sample is "abc" and "de".

The two chosen substrings for the third sample are "ded" and "cbacbca".

In the fourth sample, we can‘t choose such two non-empty substrings, so the answer is 0.

 

状压dp，解释在代码中注释掉了，自己看吧

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using  namespace std;
const int maxn=16;
int dp[1<<16+5];
char str[2005];
int main(){
       int t;
    scanf("%d",&t);
    while(t--){
      int n,k;
        memset(dp,0,sizeof(dp));
        scanf("%d%d",&n,&k);
        scanf("%s",str);
        for(int i=0;i<n;i++){//我们将16种字母划分为16位，直接dp[0~1<<16-1],对于每一位，如果值为0，代表那么字母不存在，如果为1
                             //代表那个字母存在，
            int t=0;
           for(int j=i;j<n;j++){
               t|=1<<(str[j]-‘a‘);
               dp[t]=max(dp[t],j-i+1);//求出某种状态的最长序列，，，
           }
        }

        int s=1<<k;
        for(int i=0;i<s;i++){
           for(int j=0;j<k;j++){
              if(i&(1<<j))//这里从少的字符遍历到多的字符，迭代累加，
                dp[i]=max(dp[i],dp[i^(1<<j)]);//每次逐位进行比较，如果i字符串中包含地j个字符，那么更新dp值
           }
        }
        int ans=0;
        for(int i=0;i<s;i++)
            ans=max(ans,dp[i]*(dp[(s-1)^i]));//calculate
        printf("%d\n",ans);

    }
    return 0;
}

 

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