【codeforces】2014-2015 ACM-ICPC, Central Europe Regional Contest (CERC 14) F Vocabulary【dp】

传送门：【codeforces】2014-2015 ACM-ICPC, Central Europe Regional Contest (CERC 14) F Vocabulary

题目分析：蒽。。。。这个dp嘛其实也是满简单易懂的，就是要考虑的地方略多一点。。。

设字典序最小的为a串，中间的为b串，最大的为c串。

首先我们预处理出'?'个数在前i个字符位置的个数num[i]（三个字符串的都算上）。

然后我们分三种情况讨论：

1.第i个位置之前三个串的字典序相同（每个位置三串的字母都相同），第i+1个位置a[i + 1] < b[i + 1] < c[i + 1]。

2.第i个位置之前a、b串的字典序相同，且b串的字典序小于c串，第i+1个位置a[i + 1] < b[i + 1]。

3.第i个位置之前b、c串的字典序相同，且a串的字典序小于b串，第i+1个位置b[i + 1] < c[i + 1]。

当第i个位置三个字符串的大小可以区分时，后面的'?'都可以随便取了，所以要乘上26^(num[n-1]-num[i])，其中n为三个字符串长度的最大值，字符串下标从0开始。

对于每一种情况，第i个位置开始区分字典序时，就是ans+=dp[x][i-1]*(当前的情况数)*(26^(num[n-1]-num[i]))。

dp[x][i - 1]*(当前的情况数)表示情况x在第i个位置符合情况x的定义的方案数。

剩下的就是无比繁琐的if了。。。。（对我来说是这样。。整个人都不好了，写了两天都要写晕了。。。各种细节问题）

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

typedef long long LL ;

#define rep( i , a , b ) for ( int i = ( a ) ; i <  ( b ) ; ++ i )
#define For( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define rev( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define clr( a , x ) memset ( a , x , sizeof a )

const int MAXN = 1000005 ;
const int mod = 1e9 + 9 ;

int pow[MAXN * 3] ;
int num[MAXN] ;
char a[MAXN] ;
char b[MAXN] ;
char c[MAXN] ;
int na , nb , nc , n ;
LL dpa[MAXN] , dpb[MAXN] , dpc[MAXN] ;

int sigma ( int a , int b ) {
	return ( a + b ) * ( b - a + 1 ) / 2 ;
}

void solve () {
	int ans = 0 ;
	scanf ( "%s%s%s" , a , b , c ) ;
	na = strlen ( a ) ;
	nb = strlen ( b ) ;
	nc = strlen ( c ) ;
	n = max ( max ( na , nb ) , nc ) ;
	rep ( i , 0 , n ) {
		num[i] = 0 ;
		dpa[i] = 0 ;
		dpb[i] = 0 ;
		dpc[i] = 0 ;
	}
	dpa[0] = 1 ;
	rep ( i , 0 , n ) {
		if ( i < na ) {
			if ( a[i] == '?' ) ++ num[i] ;
			else a[i] -= 'a' - 1 ;
		} else a[i] = 0 ;
		if ( i < nb ) {
			if ( b[i] == '?' ) ++ num[i] ;
			else b[i] -= 'a' - 1 ;
		} else b[i] = 0 ;
		if ( i < nc ) {
			if ( c[i] == '?' ) ++ num[i] ;
			else c[i] -= 'a' - 1 ;
		} else c[i] = 0 ;
	}
	rep ( i , 1 , n ) num[i] += num[i - 1] ;
	int ans1 = 0 , ans2 = 0 , ans3 = 0 ;
	rep ( i , 0 , n ) {
		int Pow = pow[num[n - 1] - num[i]] ;
		int cnt = ( a[i] == '?' ) + ( b[i] == '?' ) + ( c[i] == '?' ) ;
		if ( cnt == 3 ) {
			ans1 = ( ans1 + dpa[i] * 2600 % mod * Pow ) % mod ;
			dpa[i + 1] = dpa[i] * 26 % mod ;
		}
		if ( cnt == 2 ) {
			if ( a[i] != '?' ) {
				ans1 = ( ans1 + dpa[i] * sigma ( 1 , 26 - a[i] - 1 ) % mod * Pow ) % mod ;
				if ( a[i] == 0 ) break ;
				dpa[i + 1] = dpa[i] ;
			}
			if ( b[i] != '?' ) {
				if ( b[i] == 0 ) break ;
				ans1 = ( ans1 + dpa[i] * ( b[i] - 1 ) * ( 26 - b[i] ) % mod * Pow ) % mod ;
				dpa[i + 1] = dpa[i] ;
			}
			if ( c[i] != '?' ) {
				if ( c[i] == 0 ) break ;
				ans1 = ( ans1 + dpa[i] * sigma ( 1 , c[i] - 2 ) % mod * Pow ) % mod ;
				dpa[i + 1] = dpa[i] ;
			}
		}
		if ( cnt == 1 ) {
			if ( a[i] == '?' ) {
				if ( b[i] == 0 || c[i] == 0 ) break ;
				if ( b[i] < c[i] ) ans1 = ( ans1 + dpa[i] * ( b[i] - 1 ) % mod * Pow ) % mod ;
				if ( b[i] != c[i] ) break ;
				dpa[i + 1] = dpa[i] ;
			}
			if ( b[i] == '?' ) {
				if ( c[i] == 0 ) break ;
				if ( a[i] < c[i] ) ans1 = ( ans1 + dpa[i] * ( c[i] - a[i] - 1 ) % mod * Pow ) % mod ;
				if ( a[i] == 0 ) break ;
				if ( a[i] != c[i] ) break ;
				dpa[i + 1] = dpa[i] ;
			}
			if ( c[i] == '?' ) {
				if ( b[i] == 0 ) break ;
				if ( a[i] < b[i] ) ans1 = ( ans1 + dpa[i] * ( 26 - b[i] ) % mod * Pow ) % mod ;
				if ( a[i] == 0 ) break ;
				if ( a[i] != b[i] ) break ;
				dpa[i + 1] = dpa[i] ;
			}
		}
		if ( cnt == 0 ) {
			if ( a[i] <  b[i] && b[i] <  c[i] ) ans1 = ( ans1 + dpa[i] * Pow ) % mod ;
			if ( a[i] != b[i] || b[i] != c[i] ) break ;
			dpa[i + 1] = dpa[i] ;
		}
	}
	rep ( i , 0 , n ) {
		int Pow = pow[num[n - 1] - num[i]] ;
		int cnt = ( a[i] == '?' ) + ( b[i] == '?' ) + ( c[i] == '?' ) ;
		if ( cnt == 3 ) {
			ans2 = ( ans2 + dpb[i] * sigma ( 1 , 25 ) * 26 % mod * Pow ) % mod ;
			dpb[i + 1] = ( dpb[i] * 26 * 26 + dpa[i] * sigma ( 1 , 25 ) ) % mod ;
		}
		if ( cnt == 2 ) {
			if ( a[i] != '?' ) {
				ans2 = ( ans2 + dpb[i] * ( 26 - a[i] ) * 26 % mod * Pow ) % mod ;
				if ( a[i] == 0 ) break ;
				dpb[i + 1] = ( dpb[i] * 26 + dpa[i] * ( 26 - a[i] ) ) % mod ;
			}
			if ( b[i] != '?' ) {
				if ( b[i] == 0 ) break ;
				ans2 = ( ans2 + dpb[i] * ( b[i] - 1 ) * 26 % mod * Pow ) % mod ;
				dpb[i + 1] = ( dpb[i] * 26 + dpa[i] * ( 26 - b[i] ) ) % mod ;
			}
			if ( c[i] != '?' ) {
				ans2 = ( ans2 + dpb[i] * sigma ( 1 , 25 ) % mod * Pow ) % mod ;
				if ( c[i] > 1 ) dpb[i + 1] = ( dpb[i] * 26 + dpa[i] * ( c[i] - 1 ) ) % mod ;
				else dpb[i + 1] = dpb[i] * 26 % mod ;
			}
		}
		if ( cnt == 1 ) {
			if ( a[i] == '?' ) {
				if ( b[i] == 0 ) break ;
				ans2 = ( ans2 + dpb[i] * ( b[i] - 1 ) % mod * Pow ) % mod ;
				if ( b[i] < c[i] ) dpb[i + 1] = ( dpb[i] + dpa[i] ) % mod ;
				else dpb[i + 1] = dpb[i] ;
			}
			if ( b[i] == '?' ) {
				ans2 = ( ans2 + dpb[i] * ( 26 - a[i] ) % mod * Pow ) % mod ;
				if ( a[i] == 0 ) break ;
				if ( a[i] < c[i] ) dpb[i + 1] = ( dpb[i] + dpa[i] ) % mod ;
				else dpb[i + 1] = dpb[i] ;
			}
			if ( c[i] == '?' ) {
				if ( a[i] <  b[i] ) ans2 = ( ans2 + dpb[i] * 26 % mod * Pow ) % mod ;
				if ( a[i] != b[i] ) break ;
				dpb[i + 1] = ( dpb[i] * 26 + dpa[i] * ( 26 - b[i] ) ) % mod ;
			}
		}
		if ( cnt == 0 ) {
			if ( a[i] <  b[i] ) ans2 = ( ans2 + dpb[i] * Pow ) % mod ;
			if ( a[i] != b[i] ) break ;
			if ( b[i] <  c[i] ) dpb[i + 1] = ( dpb[i] + dpa[i] ) % mod ;
			else dpb[i + 1] = dpb[i] ;
		}
	}
	rep ( i , 0 , n ) {
		int Pow = pow[num[n - 1] - num[i]] ;
		int cnt = ( a[i] == '?' ) + ( b[i] == '?' ) + ( c[i] == '?' ) ;
		if ( cnt == 3 ) {
			ans3 = ( ans3 + dpc[i] * sigma ( 1 , 25 ) * 26 % mod * Pow ) % mod ;
			dpc[i + 1] = ( dpc[i] * 26 * 26 + dpa[i] * sigma ( 1 , 25 ) ) % mod ;
		}
		if ( cnt == 2 ) {
			if ( a[i] != '?' ) {
				ans3 = ( ans3 + dpc[i] * sigma ( 1 , 25 ) % mod * Pow ) % mod ;
				dpc[i + 1] = ( dpc[i] * 26 + dpa[i] * ( 26 - a[i] ) ) % mod ;
			}
			if ( b[i] != '?' ) {
				ans3 = ( ans3 + dpc[i] * 26 * ( 26 - b[i] ) % mod * Pow ) % mod ;
				if ( b[i] == 0 ) break ;
				dpc[i + 1] = ( dpc[i] * 26 + dpa[i] * ( b[i] - 1 ) ) % mod ;
			}
			if ( c[i] != '?' ) {
				if ( c[i] == 0 ) break ;
				ans3 = ( ans3 + dpc[i] * 26 * ( c[i] - 1 ) % mod * Pow ) % mod ;
				dpc[i + 1] = ( dpc[i] * 26 + dpa[i] * ( c[i] - 1 ) ) % mod ;
			}
		}
		if ( cnt == 1 ) {
			if ( a[i] == '?' ) {
				if ( b[i] < c[i] ) ans3 = ( ans3 + dpc[i] * 26 % mod * Pow ) % mod ;
				if ( b[i] != c[i] ) break ;
				dpc[i + 1] = ( dpc[i] * 26 + dpa[i] * ( b[i] - 1 ) ) % mod ;
			}
			if ( b[i] == '?' ) {
				if ( c[i] == 0 ) break ;
				ans3 = ( ans3 + dpc[i] * ( c[i] - 1 ) % mod * Pow ) % mod ;
				if ( a[i] < c[i] ) dpc[i + 1] = ( dpc[i] + dpa[i] ) % mod ;
				else dpc[i + 1] = dpc[i] ;
			}
			if ( c[i] == '?' ) {
				ans3 = ( ans3 + dpc[i] * ( 26 - b[i] ) % mod * Pow ) % mod ;
				if ( b[i] == 0 ) break ;
				if ( a[i] < b[i] ) dpc[i + 1] = ( dpc[i] + dpa[i] ) % mod ;
				else dpc[i + 1] = dpc[i] ;
			}
		}
		if ( cnt == 0 ) {
			if ( b[i] <  c[i] ) ans3 = ( ans3 + dpc[i] * Pow ) % mod ;
			if ( b[i] != c[i] ) break ;
			if ( a[i] <  b[i] ) dpc[i + 1] = ( dpc[i] + dpa[i] ) % mod ;
			else dpc[i + 1] = dpc[i] ;
		}
	}
	/*rep ( i , 0 , n ) printf ( "%d " , dpa[i] ) ;
	printf ( "\n" ) ;
	rep ( i , 0 , n ) printf ( "%d " , dpb[i] ) ;
	printf ( "\n" ) ;
	rep ( i , 0 , n ) printf ( "%d " , dpc[i] ) ;
	printf ( "\n" ) ;
	printf ( "%d %d %d\n" , ans1 , ans2 , ans3 ) ;*/
	ans = ( ( LL ) ans1 + ans2 + ans3 ) % mod ;
	printf ( "%d\n" , ans ) ;
}

int main () {
	//freopen ( "1.in" , "r" , stdin ) ;
	//freopen ( "12.out" , "w" , stdout ) ;
	int T ;
	scanf ( "%d" , &T ) ;
	pow[0] = 1 ;
	For ( i , 1 , 3000000 ) pow[i] = pow[i - 1] * 26LL % mod ;
	while ( T -- ) solve () ;
	return 0 ;
}