CodeForces - 1131 E String Multiplication【字符串】
E. String Multiplication
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Roman and Denis are on the trip to the programming competition. Since the trip was long, they soon got bored, and hence decided to came up with something. Roman invented a pizza's recipe, while Denis invented a string multiplication. According to Denis, the result of multiplication (product) of strings ss of length mm and tt is a string t+s1+t+s2+…+t+sm+tt+s1+t+s2+…+t+sm+t, where sisi denotes the ii-th symbol of the string ss, and "+" denotes string concatenation. For example, the product of strings "abc" and "de" is a string "deadebdecde", while the product of the strings "ab" and "z" is a string "zazbz". Note, that unlike the numbers multiplication, the product of strings ss and tt is not necessarily equal to product of tt and ss.
Roman was jealous of Denis, since he invented such a cool operation, and hence decided to invent something string-related too. Since Roman is beauty-lover, he decided to define the beauty of the string as the length of the longest substring, consisting of only one letter. For example, the beauty of the string "xayyaaabca" is equal to 33, since there is a substring "aaa", while the beauty of the string "qwerqwer" is equal to 11, since all neighboring symbols in it are different.
In order to entertain Roman, Denis wrote down nn strings p1,p2,p3,…,pnp1,p2,p3,…,pn on the paper and asked him to calculate the beauty of the string (…(((p1⋅p2)⋅p3)⋅…)⋅pn(…(((p1⋅p2)⋅p3)⋅…)⋅pn, where s⋅ts⋅t denotes a multiplication of strings ss and tt. Roman hasn't fully realized how Denis's multiplication works, so he asked you for a help. Denis knows, that Roman is very impressionable, he guarantees, that the beauty of the resulting string is at most 109109.
Input
The first line contains a single integer nn (2≤n≤1000002≤n≤100000) — the number of strings, wroted by Denis.
Next nn lines contain non-empty strings p1,p2,…,pnp1,p2,…,pn, consisting of lowercase english letters.
It's guaranteed, that the total length of the strings pipi is at most 100000100000, and that's the beauty of the resulting product is at most 109109.
Output
Print exactly one integer — the beauty of the product of the strings.
Examples
input
Copy
3 a b a
output
Copy
3
input
Copy
2 bnn a
output
Copy
1
Note
In the first example, the product of strings is equal to "abaaaba".
In the second example, the product of strings is equal to "abanana".
题意：按照题目叙述规则，将N个字符串计算后求最长的连续相等的子串。
根据规则，当两个字符串s，t相乘时，用dp[i][30]表示第i个s中每个字符的最长连续相等的长度
（1）当t中子串全部都一样时，dp[i][j]=t.size()*(dp[i-1][j]+1)+dp[i-1][j]
（2）当t中存在不一样的子串时，dp[i][j] = t的前缀+t的后缀+1
注意一下，最长的长度也可以由t串提供。
最后求dp[n][i]的最大值即可。
#include "bits/stdc++.h"
using namespace std;
int dp[100004][30];
int main()
{
int n;
scanf("%d",&n);
string s;
for (int i = 1; i <= n; ++i) {
cin>>s;
int tlen[30]={0},maxlen[30]={0};
for (int j = 0; j < s.size(); ++j) {//求t中最长连续相等子串长度
if(j&&s[j]==s[j-1])tlen[s[j]-'a']++;
else tlen[s[j]-'a']=1;
maxlen[s[j]-'a']=max(maxlen[s[j]-'a'],tlen[s[j]-'a']);
}
for (int j = 0; j < 26; ++j) {
dp[i][j]=maxlen[j];
int l=0,r=s.size()-1;
while(s[l]==j+'a'&&l<=r)l++;//求前缀
while(s[r]==j+'a'&&l