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codeforces 766 C Mahmoud and a Message [基礎(chǔ)DP]【動(dòng)態(tài)規(guī)劃】

2019-11-10 16:59:20
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題目連接:http://codeforces.com/contest/766/PRoblem/C

———————————————————————–. 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(chǎn)’ 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(chǎn)’ appears in a substring of length 3, while a1?=?2. ———————————————————————–. 題目大意: 一個(gè)長(zhǎng)度為n的僅含26小字母的字符串,要將其按要求分割成幾段,其中26種字母只能在長(zhǎng)度為Ax的字串中,問你劃分的種類數(shù)是多少,字串的長(zhǎng)度最長(zhǎng)是多少,劃分出來的字符串個(gè)數(shù)最少是多少,

解題思路: 還是dp

對(duì)于子串中以第i個(gè)字符為止時(shí),我們可以將其劃分到 ∈[i?Ai+1,i]的位置上,我們從這個(gè)位置向前尋找,一次次判斷就行了..

同時(shí)就能維護(hù)字串的最長(zhǎng)長(zhǎng)度了

字串的最少個(gè)數(shù)在一次次向前找中,維護(hù)最小值即可.

詳見代碼及注釋

附本題代碼 ———————————————————————–.

int a[30];int dp[3][1010];/**dp[0][i] 截止到第i個(gè)字符為止的劃分方案數(shù) dp[1][i] 一個(gè)個(gè)的維護(hù)可劃分的最大長(zhǎng)度 ,dp[2][i] 前一個(gè)的劃分點(diǎn)轉(zhuǎn)移到i,值+1, (劃分個(gè)數(shù)最少 自然是長(zhǎng)度越長(zhǎng)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?yàn)锳_j有它的范圍 ,不能超過 dp[0][i] = (dp[0][i] + dp[0][j]) % MOD; //劃分方案數(shù), dp[1][i] = max(dp[1][i], max(i - j, dp[1][j]));//就是維護(hù)最大值 dp[2][i] = min(dp[2][i], dp[2][j] + 1); //最少的劃分個(gè)數(shù), } } cout << dp[0][n] << endl; cout << dp[1][n] << endl; cout << dp[2][n] << endl; return 0;}
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