[原创]codeforces 766 C Mahmoud and a Message [基础DP]【动态规划】

2017-02-08 22:42:09 Tabris_ 阅读数：499

博客爬取于2020-06-14 22:41:47
以下为正文

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https://blog.csdn.net/qq_33184171/article/details/54934055

题目连接：http://codeforces.com/contest/766/problem/C

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C. Mahmoud and a Message
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Mahmoud wrote a message s of length n. He wants to send it as a birthday present to his friend Moaz who likes strings. He wrote it on a magical paper but he was surprised because some characters disappeared while writing the string. That’s because this magical paper doesn’t allow character number i in the English alphabet to be written on it in a string of length more than ai. For example, if a1 = 2 he can’t write character ‘a’ on this paper in a string of length 3 or more. String “aa” is allowed while string “aaa” is not.

Mahmoud decided to split the message into some non-empty substrings so that he can write every substring on an independent magical paper and fulfill the condition. The sum of their lengths should be n and they shouldn’t overlap. For example, if a1 = 2 and he wants to send string “aaa”, he can split it into “a” and “aa” and use 2 magical papers, or into “a”, “a” and “a” and use 3 magical papers. He can’t split it into “aa” and “aa” because the sum of their lengths is greater than n. He can split the message into single string if it fulfills the conditions.

A substring of string s is a string that consists of some consecutive characters from string s, strings “ab”, “abc” and “b” are substrings of string “abc”, while strings “acb” and “ac” are not. Any string is a substring of itself.

While Mahmoud was thinking of how to split the message, Ehab told him that there are many ways to split it. After that Mahmoud asked you three questions:

How many ways are there to split the string into substrings such that every substring fulfills the condition of the magical paper, the sum of their lengths is n and they don’t overlap? Compute the answer modulo 109 + 7.
What is the maximum length of a substring that can appear in some valid splitting?
What is the minimum number of substrings the message can be spit in?
Two ways are considered different, if the sets of split positions differ. For example, splitting “aa|a” and “a|aa” are considered different splittings of message “aaa”.

Input
The first line contains an integer n (1 ≤ n ≤ 103) denoting the length of the message.

The second line contains the message s of length n that consists of lowercase English letters.

The third line contains 26 integers a1, a2, …, a26 (1 ≤ ax ≤ 103) — the maximum lengths of substring each letter can appear in.

Output
Print three lines.

In the first line print the number of ways to split the message into substrings and fulfill the conditions mentioned in the problem modulo 109 + 7.

In the second line print the length of the longest substring over all the ways.

In the third line print the minimum number of substrings over all the ways.

Examples
input
3
aab
2 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
output
3
2
2
input
10
abcdeabcde
5 5 5 5 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
output
401
4
3
Note
In the first example the three ways to split the message are:

a|a|b
aa|b
a|ab
The longest substrings are “aa” and “ab” of length 2.

The minimum number of substrings is 2 in “a|ab” or “aa|b”.

Notice that “aab” is not a possible splitting because the letter ‘a’ appears in a substring of length 3, while a1 = 2.
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题目大意:
一个长度为n的仅含26小字母的字符串,要将其按要求分割成几段,其中26种字母只能在长度为Ax的字串中,问你划分的种类数是多少,字串的长度最长是多少,划分出来的字符串个数最少是多少,

解题思路:
还是dp

对于子串中以第i个字符为止时,我们可以将其划分到 $ \in \left[ i-A_i+1,i \right]$的位置上,我们从这个位置向前寻找,一次次判断就行了…

同时就能维护字串的最长长度了

字串的最少个数在一次次向前找中,维护最小值即可.

详见代码及注释

附本题代码
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int a[30];
int dp[3][1010];
/**
dp[0][i]  截止到第i个字符为止的划分方案数
dp[1][i]  一个个的维护可划分的最大长度   ，
dp[2][i]  前一个的划分点转移到i，值+1，  （划分个数最少 自然是长度越长越好）
*/
string str;
int main(){
    int n;
    cin>>n;
    cin>>str;
    for(int i=0;i<26;i++) cin>>a[i];

    dp[0][0] = 1;
    for(int i = 1; i <= n; ++i){
        int len  =  INF;
        dp[1][i] = -INF;
        dp[2][i] =  INF;
        for(int j = i - 1; j >= 0; --j){
            len = min(len, a[str[j] - 'a']);
            if(len < i - j) break;                         //因为A_j有它的范围 ,不能超过
            dp[0][i] = (dp[0][i] + dp[0][j]) % MOD;        //划分方案数，
            dp[1][i] = max(dp[1][i], max(i - j, dp[1][j]));//就是维护最大值
            dp[2][i] = min(dp[2][i], dp[2][j] + 1);        //最少的划分个数，
        }
    }

    cout << dp[0][n] << endl;
    cout << dp[1][n] << endl;
    cout << dp[2][n] << endl;

    return 0;
}