link:http://acm.hdu.edu.cn/showproblem.php?pid=4632 refer to: o(╯□╰)o……明明百度找的题解,然后后来就找不到我看的那份了,这位哥们对不住了…… #include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <cctype> #i…

题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=4632 题意:求回文串子串的的个数. 思路:看转移方程就能理解了. dp[i][j] 表示区间i j 之间的回文子串的个数. 状态转移方程:dp[i][j]=dp[i][j-1]+dp[i+1][j]-dp[i+1][j-1];         if(str[i]==str[j]) dp[i][j]=dp[i][j]+dp[i+1][j-1]+1; 代码: #include<iostrea…

Palindrome subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65535 K (Java/Others)Total Submission(s): 4513    Accepted Submission(s): 1935 Problem Description In mathematics, a subsequence is a sequence that can be derived f…

Palindrome subsequence Problem Description In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence <A, B, D> is a…

标签(空格分隔): 区间qp Palindrome subsequence \[求一个string的 回文子序列 的个数 \] 少废话,上代码. #include<bits/stdc++.h> //头文件没有什么问题 using namespace std; //名字空间没有什么问题 const int maxn = 1000 +5; // 字符串的最大长度 const int mod = 10007;//题目规定的模数 int n,cnt;//n:有多少个字符串,cnt:记录当前有多少个ca…

Problem Description In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence <A, B, D> is a subsequence of <A,…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4632 题意:给你一个序列,问你该序列中有多少个回文串子序列,可以不连续. 思路:dp[i][j]表示序列i到j中具有的回文串序列数. 当s[i]==s[j], dp[i][j]=dp[i+1][j]+dp[i][j-1]+1 否则 dp[i][j]=dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]: #include <iostream> #include <cstdio…

题目 参考自博客:http://blog.csdn.net/u011498819/article/details/38356675 题意:查找这样的子回文字符串(未必连续,但是有从左向右的顺序)个数. 简单的区间dp,哎,以为很神奇的东西,其实也是dp,只是参数改为区间,没做过此类型的题,想不到用dp,以后就 知道了,若已经知道[0,i],推[0,i+1], 显然还要从i+1 处往回找,dp方程也简单: dp[j][i]=(dp[j+1][i]+dp[j][i-1]+10007-dp[j+1][…

题意 给定一个字符串,问有多少个回文子串(两个子串可以一样). 思路 注意到任意一个回文子序列收尾两个字符一定是相同的,于是可以区间dp,用dp[i][j]表示原字符串中[i,j]位置中出现的回文子序列的个数,有递推关系: dp[i][j]=dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]  (*) 如果i和j位置出现的字符相同,那么dp[i][j]可以由dp[i+1][j-1]中的子序列加上这两个字符构成回文子序列,也就是 dp[i][j]+=dp[i+1][j-1],注意…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4632 问题要求回答一串字符串中回文子序列的数量,例如acbca就有 a,c,b,c,a,cc,aa,aca,aca(注意这两个aca的c是不同位置的c,都要累计),aba,cbc,acca,acbca.共13种. 我们如果构造dp[i][j]为区间从i-j的回文子序列个数,当i==j时dp[i][j]=1,当i!=j时,如果字符串i,j位相等,他们便可以从dp[i+1,j-1]转移而来,即dp[i]…

题目链接 题目大意 给你几个字符串 (1<len(s)<1000) ,要你求每个字符串的回文序列个数.对于10008取模. Solution 区间DP. 比较典型的例题. 状态定义: 令 \(f[i][j]\) 表示 \(i\) 到 \(j\) 的回文序列个数,\(s\) 为给出的字符串. 状态转移: \(s[i]\neq s[j]\) 那么此时 \(f[i][j]\) 即为\(f[i][j-1]\),\(f[i+1][j]\)之和. 但由于 \(i+1->j-1\)的我们明显重复统计了…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4632 题目大意:给你若干个字符串,回答每个字符串有多少个回文子序列(可以不连续的子串).解题思路: 设dp[i][j]为[i,j]的回文子序列数,那么得到状态转移方程: dp[i][j]=(dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]+MOD)%MOD if(str[i]==str[j]) dp[i][j]+=dp[i-1][j+1]+1 代码: #include<cstdi…

Palindrome subsequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65535 K (Java/Others) Total Submission(s): 2836 Accepted Submission(s): 1160 Problem Description In mathematics, a subsequence is a sequence that can be derived from a…

Palindrome subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65535 K (Java/Others)Total Submission(s): 558    Accepted Submission(s): 203 Problem Description In mathematics, a subsequence is a sequence that can be derived fro…

Palindrome subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65535 K (Java/Others) Total Submission(s): 2595    Accepted Submission(s): 1039 Problem Description In mathematics, a subsequence is a sequence that can be derived…

Palindrome subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65535 K (Java/Others)Total Submission(s): 2232    Accepted Submission(s): 889 Problem Description In mathematics, a subsequence is a sequence that can be derived fr…

Palindrome subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65535 K (Java/Others)Total Submission(s): 88    Accepted Submission(s): 26 Problem Description In mathematics, a subsequence is a sequence that can be derived from…

废话不多说,直接上题: 1580:加分二叉树 时间限制: 1000 ms         内存限制: 524288 KB提交数: 121     通过数: 91 [题目描述] 原题来自:NOIP 2003 设一个 n 个节点的二叉树 tree 的中序遍历为 (1,2,3,⋯,n),其中数字 1,2,3,⋯,n 为节点编号.每个节点都有一个分数(均为正整数),记第 i 个节点的分数为 di ,tree 及它的每个子树都有一个加分,任一棵子树 subtree(也包含 tree 本身)的加分计算方法如…

废话不多说,直接上题:  P4170 [CQOI2007]涂色 题目描述 假设你有一条长度为5的木版,初始时没有涂过任何颜色.你希望把它的5个单位长度分别涂上红.绿.蓝.绿.红色,用一个长度为5的字符串表示这个目标:RGBGR. 每次你可以把一段连续的木版涂成一个给定的颜色,后涂的颜色覆盖先涂的颜色.例如第一次把木版涂成RRRRR,第二次涂成RGGGR,第三次涂成RGBGR,达到目标. 用尽量少的涂色次数达到目标. 输入格式 输入仅一行,包含一个长度为n的字符串,即涂色目标.字符串中的每个字符都…

In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence <A, B, D> is a subsequence of <A, B, C, D, E, F>. (…

题意1:问你一个串有几个不连续子序列(相同字母不同位置视为两个) 题意2:问你一个串有几种不连续子序列(相同字母不同位置视为一个,空串视为一个子序列) 思路1:由容斥可知当两个边界字母相同时 dp[i][j] = dp[i + 1][j] + dp[i][j - 1] - dp[i + 1][j - 1] + dp[i + 1][j - 1] + 1;当两个字母不同时 dp[i][j] = dp[i + 1][j] + dp[i][j - 1] - dp[i + 1][j - 1].然后区间DP…

个人心得:动态规划真的是够烦人的,这题好不容易写出了转移方程,结果超时,然后看题解,为什么这些题目都是这样一步一步的 递推,在我看来就是懵逼的状态,还有那个背包也是,硬是从最大的V一直到0,而这个就是从把间距为1到ch.size()全部算出来,难道 这就是动态规划,无后效性,即每一步都是最优的状态,所以把所有状况全部解决然后就可以一步一步往后面推了??值得深思 网上题解: 分析:我们知道求添加最少的字母让其回文是经典dp问题,转化成LCS求解.这个是一个很明显的区间dp 我们定义dp [ i ]…

Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 18765    Accepted Submission(s): 7946 Problem Description A subsequence of a given sequence is the given sequence with some ele…

A palindrome is a nonempty string over some alphabet that reads the same forwardand backward. Examples of palindromes are all strings of length 1, civic,racecar, and aibohphobia (fear of palindromes).Give an efficient algorithm to find the longest pa…

Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 55301    Accepted Submission(s): 25537 Problem Description A subsequence of a given sequence is the given sequence with some el…

Cheapest Palindrome Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7148   Accepted: 3452 Description Keeping track of all the cows can be a tricky task so Farmer John has installed a system to automate it. He has installed on each cow a…

有n个混合物排成一排,每个混合物有一个颜色值0<=color<=99, 规定合并只能合并相邻两个, 将颜色a的混合物与颜色b的混合物合并后,颜色为( a+b ) % 100,并产生a*b的污染, 现在要将所有混合物合并,问产生污染的最小值. [区间动规]很经典的区间动规 dp[i][j] = min { dp[i][k] + dp[k+1][j] + sum[i][k]*sum[k+1][j] } 具体的DP次序详见代码: #include<cstdio> #include<…

题目链接:http://poj.org/problem?id=1458 思路分析:经典的最长公共子序列问题(longest-common-subsequence proble),使用动态规划解题. 1)问题定义:给定两个序列X=<X1, X2, ...., Xm>和Y = <Y1, Y2, ...., Yn>,要求求出X和Y长度最长的最长公共子序列: 2)问题分析: <1>动态规划问题都是多阶段决策最优化问题:在这些问题中,问题可以被划分为多个阶段,每个阶段都需要作出一…

题目链接 题意:给出一些区间,求选k个区间能覆盖的最多点的数量 思路:定义dp[i][j]为前i个点取j个区间的最大值.dp[i][j]可以转移到dp[i+1][j+1]和以i+1为起点的区间终点 具体可以看代码 #include <iostream> #include <cstdio> using namespace std; #define ll long long const int maxn = 2005; int t[maxn],dp[maxn][maxn]; int m…

原文链接https://www.cnblogs.com/zhouzhendong/p/AGC026E.html 题目传送门 - AGC026E 题意 给定一个长度为 $2n$ 的字符串,包含 $n$ 个 $'a'$ 和 $n$ 个 $'b'$ . 现在,让你按照原顺序取出一些字符,按照原顺序组成新的字符串,输出所有满足条件的字符串中字典序最大的?(字典序: $'b'>'a'>'\ '$) 条件限制:当且仅当取了原序列的第 $i$ 个 $'a'$ 时,原序列的第 $i$ 个 $'b'$ 也被取了…