【CF245H】Queries for Number of Palindromes(回文树) 【CF245H】Queries for Number of Palindromes(回文树) 题面 洛谷 题解 回文树,很类似原来一道后缀自动机的题目 后缀自动机那道题 看到\(n\)的范围很小,但是\(Query\)很多 所以提前预处理出每一段\(l,r\)的答案 时间复杂度\(O(n^2+Q)\).....
intlen =strlen(s); memset(Is ,false,sizeof(Is)); for(inti = 0 ; i < len ; i++) { Is[i][i] =true; } for(intl = 1 ; l < len ; l++) { for(inti = 0 ; i < len && i + l < len ; i++) { intj = i + l; if(s[i] == s[j]) { if(l == 1) { Is[...
each query is described by two integers li, ri (1 ≤ li ≤ ri ≤ |s|). The answer to the query is the number of substrings of string s[li… ri], which are palindromes.
memset(dp,0,sizeof(dp)); memset(flag,0,sizeof(flag)); for(int i=1; i<=n; i++) { solve(i,i); solve(i,i+1); } for(int i=1; i<=n; i++) dp[i][i]=1; for(int i=n; i>=1; i--) for(int j=i+1; j<=n; j++) dp[i][j]=dp[i][j-1]+dp[i+1][j]-...
codeforces 245H 题目大意: 给出一个字符串,询问任意区间内的回文子串的个数。 题目分析: 定义isPar[i][j]表示区间字符串[i,j]是否是回文,可以通过isPar[i+1][j-1]递推得到。 定义dp[i][j]表示及区间[i,j]内的回文子串的个数,转移方程如下: dp[i][j]=dp[i+1][j]+dp[i][j−1]−dp[...
【CodeForces - 245H 】Queries for Number of Palindromes (带容斥的区间dp),题干:You'vegotastring s = s1s2... s|s| oflength |s|,consistingoflowercaseEnglishletters.Thereal
7 years ago,#^| 0 constraints on the string length in this question is 100000, so O(|S|2) won't work. →Reply Laiu 7 years ago,#| -15 Hint:The number of distinct palindromes in a string of length n can't exceed n →Reply...
I am learning this relatively new data structure with a Kattis problem called Palindromes. In this problem, we are given a string and multiple [l, r] segments and must find the number of palindromic substrings in each segment. The number of queries and length of the string are all 10^5...
题目链接:http://codeforces.com/problemset/problem/245/H 题意: 给你一个字符串s。 然后有t个询问,每个询问给出x,y,问你区间[x,y]中的回文子串的个数。 题解: 表示状态: dp[x][y] = numbers 表示区间[x,y]中的回文子串个数
题意:给定一个字符串,求区间[l,r]内有多少回文子串。 解:回文杀我。区间dp++ 首先这玩意不能像前缀和一样[l,r]=[1,r]-[1,l-1]。 那就直接设dp[i][j]为[i,j]中回文子串的数量。 考虑拿小的拼大的。如果两端不等, dp[i][j]=dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]; ...