ACM_Codeforces Round #396(Div.2)_C_Mahmoud and a Message

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 thana_i. For example, ifa₁ = 2 he can't write character 'a' on this paper in a string of length3 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 ben and they shouldn't overlap. For example, ifa₁ = 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 use3 magical papers. He can't split it into "aa" and "aa" because the sum of their lengths is greater thann. 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 strings, 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:

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".

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 integersa₁, a₂, ..., a₂₆ (1 ≤ a_x ≤ 10³) — the maximum lengths of substring each letter can appear in.

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 modulo10⁹  +  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.

3aab2 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1Output
322Input
10abcdeabcde5 5 5 5 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1Output
40143
题目解释
给定一个字符串，对每个字母进行限制（只能出现在长度小于该字母限制的字符串中）。
在该限制下，将字符串分成若干个子串。
输出该分配过程中，对字符串的分配种数，分配出的最长子串长度，分配出的最少子串数
题目分析
简单dp
对字符串从长度为0时分析,dp[0]=0
当加入第i个字符时，对从s[i]往前长度为j（1~i）的字符串依次进行判定子串（s[i-j+1],s[i])是否能构成合法子串，
如果能，则有以下状态转移方程
dp[i]+=dp[i-j];//对字符串的分配种数
f[i]=min(f[i-j],f[i]);//分配出的最少子串数
MAX=max(MAX,j);//分配出的最长子串长度
AC代码
#include<iostream>#include<string>#include<queue>#include<algorithm>using namespace std;int n, m;const int MAXN = 1e3 + 7;const int mod = 1e9 + 7;int limit[30];//每个字母使用限制long long dp[MAXN];//每个字符串的最多匹配方式int f[MAXN];//分配出的最少字串数char s[MAXN];//字符串int check(int i, int j)//检查序列是否合法{	int l = j - i + 1;//i~j之间字符数，包含i，j	for (int k = i; k <= j; ++k)	{		if (limit[s[k] - 'a']<l)return 0;	}	return 1;}int main(){	int i, j;	while (cin >> n) {		cin >> s + 1;		for (i = 0; i < 26; ++i)			cin >> limit[i];		int MAX = 0;		dp[0] = 1;		for (i = 1; i <= n; ++i)//长度为i		{			f[i] = 1e9;			for (j = 1; j <= i; ++j)//长度			{				if (check(i - j + 1, i))//后面的当前序列合法				{					dp[i] = (dp[i] + dp[i - j]) % mod;//前i个字符分开的方法数					f[i] = min(f[i], f[i - j] + 1);//最小字串数					MAX = max(MAX, j);//最长字节				}			}		}		cout << dp[n] << endl;		cout << MAX << endl;		cout << f[n] << endl;	}	return 0;}
本文借鉴于http://blog.csdn.net/QQ_33362864/article/details/54925541
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