【bzoj2272】[Usaco2011 Feb]Cowlphabet 奶牛文字 dp

题目描述

Like all bovines, Farmer John's cows speak the peculiar 'Cow'language. Like so many languages, each word in this language comprisesa sequence of upper and lowercase letters (A-Z and a-z).  A wordis valid if and only if each ordered pair of adjacent letters inthe word is a valid pair.

Farmer John, ever worried that his cows are plotting against him,recently tried to eavesdrop on their conversation. He overheard one word before the cows noticed his presence. The Cow language isspoken so quickly, and its sounds are so strange, that all that Farmer John was able to perceive was the total number of uppercaseletters, U (1 <= U <= 250) and the total number of lowercaseletters, L (1 <= L <= 250) in the word.

Farmer John knows all P (1 <= P <= 200) valid ordered pairs of adjacent letters.  He wishes to know how many different validwords are consistent with his limited data.  However, sincethis number may be very large, he only needs the value modulo97654321.

约翰的奶牛讲的是别人听不懂的“牛语”。牛语使用的字母就是英文字母，有大小写之分。牛语中存在P个合法的词素，每个词素都由两个字母组成。牛语的单词是由字母组成的序列，一个单词是有意义的充要条件是任意相邻的字母都是合法的词素。

约翰担心他的奶牛正在密谋反对他，于是最近一直在偷听他们的对话。可是，牛语太复杂了，他模模糊糊地只听到了一个单词，并确定了这个单词中有U个大写字母，L个小写字母。现在他想知道，如果这个单词是有意义的，那么有多少种可能性呢？由于这个数字太大了，你只要输出答案取97654321的余数就可以了。

输入

* Line 1: Three space-separated integers: U, L and P

* Lines 2..P+1: Two letters (each of which may be uppercase orlowercase), representing one valid ordered pair of adjacentletters in Cow.

第一行：三个用空格隔开的整数：U，L和P，1≤U．L≤250，1≤P<=200

第二行到P+1行：第i+l有两个字母，表示第i个词素，没有两个词素是完全相同的

输出

* Line 1: A single integer, the number of valid words consistent withFarmer  John's data mod 97654321.

单个整数，表示符合条件的单词数量除以97654321的余数

样例输入

2 2 7
AB
ab
BA
ba
Aa
Bb
bB

样例输出

7

---

题解

dp

f[i][j][k]表示使用i个大写字母，j个小写字母，最后一个字母是k的情况数。

注意取模。

#include <cstdio>
#define MOD 97654321
int x[210] , y[210] , f[260][260][60];
char str[5];
int getnum(char ch)
{
    return ch >= 'A' && ch <= 'Z' ? ch - 'A' : ch - 'a' + 26;
}
int main()
{
    int u , l , p , i , j , k , ans = 0;
    scanf("%d%d%d" , &u , &l , &p);
    for(i = 1 ; i <= p ; i ++ )
        scanf("%s" , str) , x[i] = getnum(str[0]) , y[i] = getnum(str[1]);
    for(i = 0 ; i < 26 ; i ++ ) f[1][0][i] = 1;
    for(i = 26 ; i < 52 ; i ++ ) f[0][1][i] = 1;
    for(i = 2 ; i <= u + l ; i ++ )
    {
        for(j = 0 ; j <= i && j <= u ; j ++ )
        {
            for(k = 1 ; k <= p ; k ++ )
            {
                if(y[k] < 26 && j > 0)
                    f[j][i - j][y[k]] = (f[j][i - j][y[k]] + f[j - 1][i - j][x[k]]) % MOD;
                if(y[k] >= 26 && i - j > 0)
                    f[j][i - j][y[k]] = (f[j][i - j][y[k]] + f[j][i - j - 1][x[k]]) % MOD;
            }
        }
    }
    for(i = 0 ; i < 52 ; i ++ )
        ans = (ans + f[u][l][i]) % MOD;
    printf("%d
" , ans);
    return 0;
}