Luogu3694 邦邦的大合唱站队 （状压DP）

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("\n\n----------\n\n")

#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;

}

Luogu3694 邦邦的大合唱站队 （状压DP）的相关教程结束。