poj 1229

dp题，神奇的替换：

ONE   匹配一个单词
ZERO_ONE  匹配零个或一个单词
ANY 可以匹配零个或多个单词

这样：
* = ONE, ANY
? = ONE, ZERO_ONE, ZERO_ONE
! = ONE, ONE, ONE, ANY

替换后即可dp。

注意这个例子：

a.b.?
!.c （输出yes）

代码：

#include<iostream>
#include<fstream>
using namespace std;

char c[400][300],c1[400][300];
int n1,n2;
char c2[120000];
int n;
bool dp[400][400];


void solve(){
	int i,j,k;
	memset(dp,0,sizeof(dp));
	dp[0][0]=1;
	for(i=1;i<=n1;i++)
		for(j=1;j<=n2;j++)
		{
			if(c[i][0]!='*'&&c[i][0]!='?'&&c[i][0]!='!'&&c1[j][0]!='*'&&c1[j][0]!='?'&&c1[j][0]!='!')
			{
				if(strcmp(c[i],c1[j])==0)
					dp[i][j]=dp[i-1][j-1];
				else
					dp[i][j]=0;
			}
			if(c[i][0]=='*')
				dp[i][j]=dp[i][j]||dp[i][j-1]||dp[i-1][j];
			if(c[i][0]=='?')
				dp[i][j]=dp[i][j]||dp[i-1][j-1];
			if(c[i][0]=='!')
				dp[i][j]=dp[i][j]||dp[i-1][j]||dp[i-1][j-1];
			if(c1[j][0]=='*')
				dp[i][j]=dp[i][j]||dp[i][j-1]||dp[i-1][j];
			if(c1[j][0]=='?')
				dp[i][j]=dp[i][j]||dp[i-1][j-1];
			if(c1[j][0]=='!')
				dp[i][j]=dp[i][j]||dp[i-1][j-1]||dp[i][j-1];
		}
		if(dp[n1][n2])
			cout<<"YES"<<endl;
		else
			cout<<"NO"<<endl;
}

void read(){
//	ifstream cin("in.txt");
	int i,j,k;
	int K;
	cin>>K;
	while(K--)
	{
		n1=n2=1;
		memset(c,0,sizeof(c));
		memset(c1,0,sizeof(c1));
		cin>>c2;
		k=0;
		for(i=0;i<strlen(c2);i++)
		{
			if(c2[i]=='.')
			{
				n1++;
				k=0;
			
			}
			else
			{
				if(c2[i]=='*')
				{
					c[n1][0]='?';
					n1++;
					c[n1][0]='*';
					
					
				}
				else
					if(c2[i]=='?')
					{
						c[n1][0]='?';
						n1++;
						c[n1][0]='!';
						n1++;
						c[n1][0]='!';
						
					}
					else
						if(c2[i]=='!')
						{
							c[n1][0]='?';
							n1++;
							c[n1][0]='?';
							n1++;
							c[n1][0]='?';
							n1++;
							c[n1][0]='*';
							
						}
						else
							c[n1][k++]=c2[i];
			}
		}
		cin>>c2;
		k=0;
		for(i=0;i<strlen(c2);i++)
		{
			if(c2[i]=='.')
			{
				n2++;
				k=0;
			
			}
			else
			{
				if(c2[i]=='*')
				{
					c1[n2][0]='?';
					n2++;
					c1[n2][0]='*';
				}
				else
					if(c2[i]=='?')
					{
						c1[n2][0]='?';
						n2++;
						c1[n2][0]='!';
						n2++;
						c1[n2][0]='!';
					}
					else
						if(c2[i]=='!')
						{
							c1[n2][0]='?';
							n2++;
							c1[n2][0]='?';
							n2++;
							c1[n2][0]='?';
							n2++;
							c1[n2][0]='*';
						}
						else
							c1[n2][k++]=c2[i];
			}
		}
		solve();
	}
}

int main(){
	read();
	return 0;
}