【bzoj1087】互不侵犯King 状态压缩dp

AC通道:http://www.lydsy.com/JudgeOnline/problem.php?id=1087

【题解】

用f[i][j][k]表示前i行放了j个棋子且第i行的状态为k的方案数。

vis[i]表示状态i是否合法,check[i][j]表示状态i,j是否可以相邻。

详见代码:

/*************
  bzoj 1087
  by chty
  2016.11.15
*************/
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
#define FILE "read"
#define up(i,j,n) for(ll i=j;i<=n;i++)
namespace INIT{
	char buf[1<<15],*fs,*ft;
	inline char getc() {return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<15,stdin),fs==ft))?0:*fs++;}
	inline ll read() {
		ll x=0,f=1;  char ch=getc();
		while(!isdigit(ch))  {if(ch=='-')  f=-1;  ch=getc();}
		while(isdigit(ch))  {x=x*10+ch-'0';  ch=getc();}
		return x*f;
	}
}using namespace INIT;
ll n,m,ans,vis[1<<9],sum[1<<9],check[1<<9][1<<9],f[11][121][1<<9];
void pre(){
	up(i,0,(1<<n)-1)if(!(i&(i<<1))){vis[i]=1; for(ll x=i;x;x>>=1)  sum[i]+=(x&1);}
	up(i,0,(1<<n)-1){
		if(!vis[i])  continue;
		up(j,0,(1<<n)-1){
			if(!vis[j]||(i&j)||(i&(j>>1))||(j&(i>>1)))  continue;
			check[i][j]=1;
		}
	}
}
int main(){
	freopen(FILE".in","r",stdin);
	freopen(FILE".out","w",stdout);
	n=read();  m=read();
	pre();  up(i,0,(1<<n)-1) if(vis[i]) f[1][sum[i]][i]=1;
	up(i,2,n)up(j,0,(1<<n)-1){
		if(!vis[j])  continue;
		up(k,0,(1<<n)-1){
			if(!vis[k]||!check[j][k])  continue;
			up(p,sum[k],m-sum[j])  f[i][p+sum[j]][j]+=f[i-1][p][k];
		}
	}
	up(i,0,(1<<n)-1)  ans+=f[n][m][i];
	printf("%lld\n",ans);
	return 0;
}


posted @ 2016-11-15 21:49  chty  阅读(197)  评论(0编辑  收藏  举报