大意: 给定串$s$, 字符集为字母表前$m$个字符, 求一个$m$排列$pos$, 使得$\sum\limits_{i=2}^n|{pos}_{s_{i-1}}-{pos}_{s_{i}}|$最小.

 

状压$dp$, 费用提前计算一下, 预处理$cost_{i,j}$表示与字符$i$相连的状态为$j$时的方案数

总复杂度是$O(n 2^n)$

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



const int N = 1e6+50;
int n, m;
char s[N];
int cnt[20][20],Log[1<<20];
int dp[1<<20], cost[20][1<<20];
void chkmin(int &a, int b) {a>b?a=b:0;}

int main() {
	printf("%d\n",n);
	scanf("%d%d%s",&n,&m,s+1);
	REP(i,2,n) {
		int a=s[i-1]-'a',b=s[i]-'a';
		++cnt[a][b],++cnt[b][a];
	}
	REP(i,0,m-1) Log[1<<i] = i;
	int mx = (1<<m)-1;
	REP(i,0,m-1) REP(j,1,mx) cost[i][j]=cost[i][j^j&-j]+cnt[i][Log[j&-j]];
	memset(dp,0x3f,sizeof dp);
	dp[0] = 0;
	REP(i,0,mx) {
		int c = 0, s = ~i&mx;
		REP(j,0,m-1) if (i>>j&1) c += cost[j][s];
		REP(j,0,m-1) if (i>>j&1^1) {
			chkmin(dp[i^1<<j],dp[i]+c+cost[j][s^1<<j]-cost[j][i]);
		}
	}
	printf("%d\n", dp[mx]);
}