hdu 4317 Unfair Nim

最新推荐文章于 2020-09-04 17:11:42 发布

最新推荐文章于 2020-09-04 17:11:42 发布 · 141 阅读

本文介绍了一种使用状态压缩动态规划解决复杂组合问题的方法，并通过一个具体示例详细讲解了实现过程，包括初始化、状态转移等关键步骤。

多校2的一道状态压缩dp，好多位运算。。。还是有点难写的

#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<cmath>
#include<cassert>
#include<cstring>
#include<iomanip>
using namespace std;
#ifdef _WIN32
#define i64 __int64
#define out64 "%I64d\n"
#define in64 "%I64d"
#else
#define i64 long long
#define out64 "%lld\n"
#define in64 "%lld"
#endif
/************ for topcoder by zz1215 *******************/
#define FOR(i,a,b)      for( int i = (a) ; i <= (b) ; i ++)
#define FFF(i,a)        for( int i = 0 ; i < (a) ; i ++)
#define FFD(i,a,b)      for( int i = (a) ; i >= (b) ; i --)
#define S64(a)          scanf(in64,&a)
#define SS(a)           scanf("%d",&a)
#define LL(a)           ((a)<<1)
#define RR(a)           (((a)<<1)+1)
#define pb              push_back
#define CL(Q)           while(!Q.empty())Q.pop()
#define MM(name,what)   memset(name,what,sizeof(name))
#define MAX(a,b)        ((a)>(b)?(a):(b))
#define MIN(a,b)        ((a)<(b)?(a):(b))
#define read            freopen("in.txt","r",stdin)
#define write           freopen("out.txt","w",stdout)

const int inf = 0x3f3f3f3f;
const i64 inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double oo = 10e9;
const double eps = 10e-9;
const double pi = acos(-1.0);
const int maxn = 1<<10;
const int maxz = 23;

int n;
int dp[maxz][maxn+1];
int a[maxz];
int s[maxz];

int find(int x)
{
	int re=0;	
	while(x)
	{
		re+=(x&1);
		x>>=1;
	}
	return re;
}


void start()
{
	for(int i=0;i<maxz;i++)
	{
		for(int j=0;j<=maxn;j++)
		{
			dp[i][j]=inf;
		}
	}
	dp[0][0]=0;
	int now,temp;
	int cost;	
	int rest;
	for(int u=1;u<maxz;u++)
	{
		for(int i=0;i<(1<<n);i++)
		{
			if(dp[u-1][i]==inf)
				continue;
			now = s[u]&i;	
			temp = s[u]|i;
			for(int k=0;k<(1<<n);k++)
			{
				if(~k&now) 
					continue;
				if(~temp&k) 
					continue;
				cost = k&~now;
				cost = find(cost);
				rest = temp&~k;
				if(find(rest)%2!=0)
				{
					if(rest == (1<<n)-1)
					{
						continue;
					}
					else
					{
						cost++;
					}				
				}
				cost*=1<<(u-1);
				dp[u][k]=min(dp[u][k],cost+dp[u-1][i]);
			}
		}
	}	
	return ;
}

int main()
{
	while(cin>>n)
	{
		MM(s,0);
		for(int i=0;i<n;i++)
		{
			cin>>a[i];		
		}	
		for(int i=1;i<maxz;i++)
		{
			for(int j=0;j<n;j++)
			{
				s[i]|=((a[j]&(1<<(i-1)))&&1)<<j;
			}			
		}
		start();
		if(dp[maxz-1][0]==inf)
		{
			cout<<"impossible"<<endl;
		}
		else
		{
			cout<<dp[maxz-1][0]<<endl;
		}
	}
	return 0;
}