ZOJ2563 之状态压缩dp

Long Dominoes Time Limit: 2 Seconds       Memory Limit: 65536 KB

Find the number of ways to tile an m*n rectangle with long dominoes -- 3*1 rectangles.

Each domino must be completely within the rectangle, dominoes must not overlap (of course, they may touch each other), each point of the rectangle must be covered.

Input

The input contains several cases. Each case stands two integers m and n (1 <= m <= 9, 1 <= n <= 30) in a single line. The input ends up with a case of m = n = 0.

Output

Output the number of ways to tile an m*n rectangle with long dominoes.

Sample Input

3 3
3 10
0 0

Sample Output

2
28

分析：由于第k行可能受第k-1,k-2行影响，所以在这里用三进制进行状态压缩,某位置o表示在该点开始竖着,1表示该点被恰好填充，2表示该点被上一行该点竖着而填充

于是判断两个状态i,j是否兼容就有了这个判断：

bool check(int i,int j){
int num=0;
while(num<m){
int a=i%3,b=j%3;
if(a == 0 && b != 1)return false;
if(a == 2 && b != 0)return false;
if(a == 1 && b == 0)return false;
if(a == 0 || a == 2)i=i/3,j=j/3,++num;
if(a == 1){
if(b == 2)i=i/3,j=j/3,++num;
else{
i=i/3,j=j/3,++num;
a=i%3,b=j%3;
if(a != 1)return false;
if(b != 1)return false;
i=i/3,j=j/3,++num;
a=i%3,b=j%3;
if(a != 1)return false;
if(b != 1)return false;
i=i/3,j=j/3,++num;
}
}
}
return true;
}

其他就是基本的状态压缩模版了

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <map>
#include <cmath>
#include <iomanip>
#define INF 99999999
typedef long long LL;
using namespace std;

const int MAX=20000+10;
int n,m;
int digit[13],bin[13];
LL dp[MAX],temp[MAX];//对应j位为0表示第j位竖着,2表示j-1位竖着,1表示j被填充

bool check(int i,int j){
	int num=0;
	while(num<m){
		int a=i%3,b=j%3;
		if(a == 0 && b != 1)return false;
		if(a == 2 && b != 0)return false;
		if(a == 1 && b == 0)return false;
		if(a == 0 || a == 2)i=i/3,j=j/3,++num;
		if(a == 1){
			if(b == 2)i=i/3,j=j/3,++num;
			else{
				i=i/3,j=j/3,++num;
				a=i%3,b=j%3;
				if(a != 1)return false;
				if(b != 1)return false;
				i=i/3,j=j/3,++num;
				a=i%3,b=j%3;
				if(a != 1)return false;
				if(b != 1)return false;
				i=i/3,j=j/3,++num;
			}
		}
	}
	return true;
}

void cal(int i){
	int size=0;
	memset(digit,0,sizeof digit);
	while(i)digit[size++]=i%3,i=i/3;
}

void dfs(int id,int k,int i,int j){//dfs搜索i能到达的状态j 
	if(k>m)return;
	if(k == m){dp[j]+=temp[i];return;}
	if(digit[k] == 0){
		dfs(id,k+1,i,j+bin[k]*2);
	}
	if(digit[k] == 2){
		dfs(id,k+1,i,j+bin[k]*1);
	}
	if(digit[k] == 1){
		dfs(id,k+1,i,j+bin[k]*0);
		if(digit[k+1] == 1 && digit[k+2] == 1)
			dfs(id,k+3,i,j+bin[k]+bin[k+1]+bin[k+2]);
	}
}

void DP(){
	int bit=1,num=1;
	for(int i=1;i<=m;++i)bit=bit*3,num=num+bit;
	num-=bit;
	memset(temp,0,sizeof temp);
	for(int i=0;i<bit;++i)if(check(i,num))temp[i]=1;
	for(int k=0;k<n-1;++k){
		for(int i=0;i<bit;++i)dp[i]=0;
		for(int i=0;i<bit;++i){
			cal(i);
			dfs(k,0,i,0);
		}
		for(int i=0;i<bit;++i)temp[i]=dp[i];
	}
	printf("%lld\n",temp[num]);
}

int main(){
	bin[0]=1;
	for(int i=1;i<=10;++i)bin[i]=bin[i-1]*3;
	while(~scanf("%d%d",&m,&n),n+m){
		DP();
	}
	return 0;
}
/*
void dfs(int id,int k,int i,int j){//由i逆推到能到达i的状态j 
	if(k>m)return;
	if(k == m){dp[i]+=temp[j];return;}
	if(digit[k] == 0){
		dfs(id,k+1,i,j+bin[k]*1);
	}
	if(digit[k] == 2){
		dfs(id,k+1,i,j+bin[k]*0);
	}
	if(digit[k] == 1){
		dfs(id,k+1,i,j+bin[k]*2);
		if(digit[k+1] == 1 && digit[k+2] == 1)
			dfs(id,k+3,i,j+bin[k]+bin[k+1]+bin[k+2]);
	}
}

void DP(){
	int bit=1,num=1;
	for(int i=1;i<=m;++i)bit=bit*3,num=num+bit;
	num-=bit;
	memset(temp,0,sizeof temp);
	for(int i=0;i<bit;++i)if(check(i,num))temp[i]=1;
	for(int k=1;k<n;++k){
		for(int i=0;i<bit;++i)dp[i]=0;
		for(int i=0;i<bit;++i){
			cal(i);
			dfs(k,0,i,0);
		}
		for(int i=0;i<bit;++i)temp[i]=dp[i];
	}
	printf("%lld\n",temp[num]);
}
*/