CF906C Party （状压DP）（未完成//鬼知道啥情况）

CF上下了数据，本地能过，一交就错

这状压不用解释，看代码就懂。

#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;

const int N = 23;

int f[1 << N];

int a[1 << N];

int q[N + 7];

int fa[1 << N];

int main(){

//FileOpen();

	int n, m;

	io >> n >> m;

	if(n == 1 || (n * (n - 1) >> 1) <= m){

		printf("0");

		return 0;

	}

	int M = (1 << n) - 1;

	R(i,0,M) f[i] = 0x3f3f3f3f; // can't memset f[] 0x7fffffff directly

	R(i,1,m){

		int u, v;

		io >> u >> v;

		a[u] |= (1 << (v - 1));

		a[v] |= (1 << (u - 1));

	}

	R(i,1,n){

		a[i] |= (1 << (i - 1));

		f[a[i]] = 1;

		q[a[i]] = i;

	}

	R(s,0,M){

		if(f[s] == 0x3f3f3f3f) continue;

		R(i,1,n){

			if(((s >> (i - 1)) & 1) && (f[s | a[i]] > (f[s] + 1))){

				f[s | a[i]] = f[s] + 1;

				q[s | a[i]] = i;

				fa[s | a[i]] = s;

			}

		}

	}

	printf("%d\n", f[M]);

	for(register int i = M; i; i = fa[i]){

		printf("%d ", q[i]);

	}

	return 0;

}

CF906C Party （状压DP）（未完成//鬼知道啥情况）的相关教程结束。