状态由(从前往后排好的长度)和(排好的团队)决定,(DP)方程挺有思考价值的。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int a = (b); a <= (c); ++ a)
#define nR(a,b,c) for(register int a = (b); a >= (c); -- a)
#define Max(a,b) ((a) > (b) ? (a) : (b))
#define Min(a,b) ((a) < (b) ? (a) : (b))
#define Fill(a,b) memset(a, b, sizeof(a))
#define Abs(a) ((a) < 0 ? -(a) : (a))
#define Swap(a,b) a^=b^=a^=b
#define ll long long
//#define ON_DEBUG
#ifdef ON_DEBUG
#define D_e_Line printf("
----------
")
#define D_e(x) cout << #x << " = " << x << endl
#define Pause() system("pause")
#define FileOpen() freopen("in.txt","r",stdin);
#else
#define D_e_Line ;
#define D_e(x) ;
#define Pause() ;
#define FileOpen() ;
#endif
struct ios{
template<typename ATP>ios& operator >> (ATP &x){
x = 0; int f = 1; char c;
for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-') f = -1;
while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
x*= f;
return *this;
}
}io;
using namespace std;
int num[21], sum[100007][21];
int bin[21];
int f[(1 << 20) + 7];
int main(){
int n, m;
io >> n >> m;
R(i,1,n){
int x;
io >> x;
R(j,1,m) sum[i][j] = sum[i - 1][j];
++num[x];
++sum[i][x];
}
bin[1] = 1;
R(i,2,m) bin[i] = bin[i - 1] << 1;
Fill(f, 0x3f3f3f3f);
f[0] = 0;
int maxx = (1 << m) - 1;
R(i,0,maxx){
int len = 0;
R(j,1,m){
if(i & bin[j]){
len += num[j];
}
}
R(j,1,m){
if(i & bin[j]){
f[i] = Min(f[i], f[i ^ bin[j]] + num[j] - (sum[len][j] - sum[len - num[j]][j]));
}
}
}
printf("%d", f[maxx]);
return 0;
}