给一个字符串,求最长回文字串有多长 #include<cstdio> #include<algorithm> #include<cstring> #define N 1000005 using namespace std; int n,m,T,p[N*2],ans; char s[2*N],t[N]; void manacher() { int id=0,pos=0,x=0; for (int i=1;i<+n;i++) { if (pos>i) x=min…

D - Palindrome Time Limit:15000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 3974 Description Andy the smart computer science student was attending an algorithms class when the professor asked the students a…

题链: http://poj.org/problem?id=3974 题解: Manacher 求最长回文串长度. 终于会了传说中的马拉车,激动.推荐一个很棒的博客:https://www.61mon.com/index.php/archives/181/ 代码: #include<cstdio> #include<cstring> #include<iostream> #define MAXN 2005000 #define filein(x) freopen(#x&…

题目链接:http://poj.org/problem?id=3974 Time Limit: 15000MS Memory Limit: 65536K Description Andy the smart computer science student was attending an algorithms class when the professor asked the students a simple question, "Can you propose an efficient…

题目链接:http://poj.org/problem?id=3974 题意:求一给定字符串最长回文子串的长度 思路:直接套模板manacher算法 code: #include <cstdio> #include <cstring> #include <algorithm> using namespace std; ; *MAXN]; *MAXN]; void Manacher(char s[], int len) { ; Ma[l++] = 'S'; Ma[l++]…

http://poj.org/problem?id=3974 模板题,Manacher算法主要利用了已匹配回文串的对称性,对前面已匹配的回文串进行利用,使时间复杂度从O(n^2)变为O(n). https://www.cnblogs.com/xiaoningmeng/p/5861154.html 详细解释 https://www.zhihu.com/question/30226229 这是复杂度O(n)的解释 代码 #include<cstdio> #include<cstring>…

Palindrome Time Limit: 15000MS   Memory Limit: 65536K Total Submissions: 12616   Accepted: 4769 Description Andy the smart computer science student was attending an algorithms class when the professor asked the students a simple question, "Can you pr…

题目链接 题意:求给定的字符串的最长回文子串 分析:做法是构造一个新的字符串是原字符串+反转后的原字符串(这样方便求两边回文的后缀的最长前缀),即newS = S + '$' + revS,枚举回文串中心位置,RMQ询问LCP = min (height[rank[l]+1] to height[rank[r]]),注意的是RMQ传入参数最好是后缀的位置,因为它们在树上的顺序未知,且左边还要+1. #include <cstdio> #include <algorithm> #in…

hash + 二分答案 数据范围肯定不能暴力,所以考虑哈希. 把前缀和后缀都哈希过之后,扫描一边字符串,对每个字符串二分枚举回文串长度,注意要分奇数和偶数 #include <iostream> #include <cstdio> #define INF 0x3f3f3f3f #define full(a, b) memset(a, b, sizeof a) using namespace std; typedef long long ll; typedef unsigned lo…

POJ 1159 Palindrome(字符串变回文:LCS) id=1159">http://poj.org/problem? id=1159 题意: 给你一个字符串, 问你做少须要在该字符串中插入几个字符能是的它变成一个回文串. 分析: 首先把原字符串和它的逆串进行匹配, 找出最长公共子序列. 那么最长公共子序列的字符串肯定是一个回文串. 所以原串剩下的部分是不构成回文的. 我们仅仅须要加入剩下部分的字符到相应位置, 原串自然就变成了一个回文. 所以本题的解为: n 减去 (原串与逆串…

http://poj.org/problem?id=3974 题意:求s的最长回文串.(|s|<=1000000) #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <iostream> using namespace std; char s[2000050]; int len[2000050], T; int main()…

http://poj.org/problem?id=3974 Manacher模板题.Menci的博客讲得很好 有一点:Menci的代码中的right我感觉是代表能延伸到的最右端点的右边的点,因为r(i)表示包括中心点的回文半径. 不知道理解有没有错误,如果神犇发现了我的理解的错误请告诉本蒟蒻qwq #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int…

[题目链接] http://poj.org/problem?id=3974 [算法] 解法1 : 字符串哈希 我们可以分别考虑奇回文子串和偶回文子串,从前往后扫描字符串,然后二分答案,检验可以用哈希 时间复杂度 : O(TNlog(N)) 解法2 Manacher算法 这个算法可以在O(n)时间内求出最长回文子串,读者可以自行查阅资料,笔者不进行详细的介绍 两种方法的效率比对            显然,Manacher算法的复杂度是优于字符串哈希的,笔者的两份代码在POJ上的运行时间分别为48…

Andy the smart computer science student was attending an algorithms class when the professor asked the students a simple question, "Can you propose an efficient algorithm to find the length of the largest palindrome in a string?" A string is sai…

题意:就是求一个串的最长回文子串....输出长度. 直接上代码吧,没什么好分析的了.   代码如下: ============================================================================================================================== #include<stdio.h> #include<string.h> #include<algorithm> us…

题目链接:http://poj.org/problem?id=3974 题意:求出给定字符串的最长回文串长度. 思路:裸的Manacher模板题. #include<iostream> #include<cstdio> #include<cstring> #include<string> #include<cmath> #include<algorithm> using namespace std; +; typedef long l…

1.链接地址: http://bailian.openjudge.cn/practice/1159/ http://poj.org/problem?id=1159 2.题目: Palindrome Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 49849   Accepted: 17153 Description A palindrome is a symmetrical string, that is, a strin…

Palindrome [题目链接]Palindrome [题目类型]最长公共子序列 &题解: 你做的操作只能是插入字符,但是你要使最后palindrome,插入了之后就相当于抵消了,所以就和在这个串中删除最少的字符,使得它回文是一样的. 那么我们可以把这个串reverse,之后的串称为s2,找s2和s的最长公共子序列就好了,因为有了LCS,接着把其他的都删掉,就是一个回文串了,因为正着读和倒着读都一样 还有POJ居然能跑5000^2 我的923MS就跑完了,还是很快的嘛,当然这题还可以滚动数组,…

A palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. You are to write a program which, given a string, determines the minimal number of characters to be inserted into the string in…

Palindrome Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 59094   Accepted: 20528 Description A palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. You are to write a…

Palindrome Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 55018   Accepted: 19024 Description A palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. You are to write a…

题目:http://poj.org/problem?id=1159 #include<iostream> #include<cstring> #include<cstdio> using namespace std; int mmin(int a,int b) { return a>b?b:a; } ][]; int main() { int n,i,j; ]; memset(d,,sizeof(d)); cin>>n; ; i<=n; i++)…

题目链接:http://poj.org/problem?id=1159 题目大意:给定一串字符,添加最少的字符,使之成为回文串. Sample Input 5 Ab3bd Sample Output 2 分析:这道题目之前有做过,就是将原字符串逆序得到另一个字符串,求它们的最长公共子序列,这样就能求得它的可以构成回文的最长字符数,用n减去即为添加最少可使之成为回文的数目. 可恨的是之前一直超内存,看了别人的解题报告,原来定义dp[MAX][MAX]时,不用int型,而是short型,内存只占in…

Description A palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. You are to write a program which, given a string, determines the minimal number of characters to be inserted into t…

题目链接:http://poj.org/problem?id=1159 思路分析:对该问题的最优子结构与最长回文子序列相同.根据最长回文子序列的状态方程稍加改变就可以得到该问题动态方程. 假设字符串为A[0, 1, ..., n],则定义状态dp[i, j]表示字符串A[i, i+1, ..., j]成为回文字符串所需要插入的最少字符数,则当A[i] == A[j+1]时, dp[i,j] = dp[i+1, j-1],否则dp[i, j] = Min(dp[i+1, j], dp[i, j+1…

Palindrome Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 51631   Accepted: 17768 Description A palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. You are to write a…

Palindrome Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 56150   Accepted: 19398 Description A palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. You are to write a…

http://poj.org/problem?id=1159 题意: 给出一个字符串,计算最少要插入多少个字符可以使得该串变为回文串. 思路: 计算出最长公共子序列,再用原长-LCS=所要添加的字符数. #include<iostream> #include<string> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; + ; char s1…

Palindrome Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 59101   Accepted: 20532 Description A palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. You are to write a…

任意门:http://poj.org/problem?id=1159 解题思路: LCS + 滚动数组 AC code: #include <cstdio> #include <iostream> #include <algorithm> #define INF 0x3f3f3f3f #define LL long long using namespace std; ; char a[MAXN], b[MAXN]; ][MAXN]; int main() { int l…