Description It is well known that a human gene can be considered as a sequence, consisting of four nucleotides, which are simply denoted by four letters, A, C, G, and T. Biologists have been interested in identifying human genes and determining their…

Description In a few months the European Currency Union will become a reality. However, to join the club, the Maastricht criteria must be fulfilled, and this is not a trivial task for the countries (maybe except for Luxembourg). To enforce that Germa…

POJ 1458 最长公共子序列 题目大意:给出两个字符串,求出这样的一 个最长的公共子序列的长度:子序列 中的每个字符都能在两个原串中找到, 而且每个字符的先后顺序和原串中的 先后顺序一致. Sample Input : abcfbc abfcab programming contest abcd mnp Sample Output 4 2 0 分析: 输入两个串s1,s2, 设dp(i,j)表示: s1的左边i个字符形成的子串,与s2左边的j个 字符形成的子串的最长公共子序列的长度(i,j从…

题意:有两个代表基因序列的字符串s1和s2,在两个基因序列中通过添加"-"来使得两个序列等长:其中每对基因匹配时会形成题中图片所示匹配值,求所能得到的总的最大匹配值. 题解:这题运用dp的解法是借用了求最长公共子序列的方法,,定义dp[i][j]代表s1以第i位结尾的串和s2以第j位结尾的串匹配时所能得到的最大匹配值:那么状态转移方程为:dp[i][j]=max( dp[i-1][j-1]+s1[i]和s2[j]的匹配值 , dp[i-1][j]+s1[i]和'-'的匹配值 , dp[…

子序列就是子序列中的元素是母序列的子集,且子序列中元素的相对顺序和母序列相同. 题目要求便是寻找两个字符串的最长公共子序列. dp[i][j]表示字符串s1左i个字符和s2左j个字符的公共子序列的最大长度. 注意s1第i个字符为s1[i-1] 于是有递推公式: 对于abcfbc和abfcab两个字符串,求公共子串的最大长度的过程如图: //#define LOCAL #include <iostream> #include <cstdio> #include <cstring…

经典的最长公共子序列问题. 状态转移方程为 : if(x[i] == Y[j]) dp[i, j] = dp[i - 1, j - 1] +1 else dp[i, j] = max(dp[i - 1], j, dp[i, j - 1]); 设有字符串X和字符串Y,dp[i, j]表示的是X的前i个字符与Y的前j个字符的最长公共子序列长度. 如果X[i] == Y[j] ,那么这个字符与之前的LCS 一定可以构成一个新的LCS: 如果X[i] != Y[j] ,则分别考察 dp[i  -1][j…

最长公共子序列可以用在下面的问题时:给你一个字符串,请问最少还需要添加多少个字符就可以让它编程一个回文串? 解法:ans=strlen(原串)-LCS(原串,反串); Sample Input abcfbc abfcab programming contest abcd mnp Sample Output 4 2 0 代码: #include <stdio.h> #include <string.h> #include <stdlib.h> #include <c…

题意: 输入俩个字符串,怎样变换使其所有字符对和最大.(字符只有'A','C','G','T','-') 其中每对字符对应的值如下: 怎样配使和最大呢. 比如: A G T G A T G -  G T T A -  G 和为 (-3)+5+5+(-2)+5+(-1) +5=14. 题解: 最长公共子序列的变形. 设dp[i][j]为a的前i个和b的前j个字符能构成的最大和. score[][]为每对字符的值,比如score['A']['G']为'A','G'这对字符对应的值. string a…

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://acm.split.hdu.edu.cn/showproblem.php?pid=1503 题意:给出两个字符串 要求输出包含两个字符串的所有字母的最短序列.注意输出的顺序不能变.//意会一下吧,我说不清=.= 思路:最长公共子序列的变形,需要记录位置.直接看代码应该就可以懂,不是很难. click here:http://www.cnblogs.com/a-clown/p/5918080.html  //hdu1159 最长公共子序列裸题. 代码: #include<i…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1243 分析:dp[i][j]表示前i个子弹去炸前j个恐怖分子得到的最大分.其实就是最长公共子序列加每个字母值为1,这里每个字母代表的值变化了一下. 状态转移方程:if(s1[i-1]==s2[j-1])dp[nxt][j]=dp[cur][j-1]+val[s1[i-1]];                              else  dp[nxt][j]=max(dp[nxt][j-1]…

求字符串和其逆的最长公共子序列,需要添加的字符数就为长度-最长公共子序列长 #include "stdio.h" #include "string.h" #define maxn 1005 char s[maxn],s1[maxn]; int dp[maxn][maxn]; int main() { ,i,j,len; scanf("%s",s); len=strlen(s); strcpy(s1,s); strrev(s1); ;i<le…

Palindrome Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 53414   Accepted: 18449 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…

看代码就懂了  不解释  3 1 1 1 1 2 2 2 1 1 1 3  第一个3 和最后一个 3 只需要一个就够了,,, #include<iostream> #include<cstring> #include<algorithm> #include<stdio.h> #include<cmath> using namespace std; ],num[],arr[]; int main( ) { int N; scanf("%d…

#include <iostream> #include <algorithm> #include <string> #include <cstring> #include <cstdio> #define MAX 1005 using namespace std; int ans[MAX][MAX]; int main(){ string s1,s2; while(cin>>s1>>s2) { memset(ans,,s…

[POJ 1080] Human Gene Functions 相似于最长公共子序列的做法 dp[i][j]表示 str1[i]相应str2[j]时的最大得分 转移方程为 dp[i][j]=max(dp[i-1][j-1]+score[str1[i]][str2[j]], max(dp[i-1][j]+score[str1[i]]['-'],dp[i][j-1]+score['-'][str2[j]]) ) 注意初始化0下标就好 代码例如以下: #include <iostream> #inc…

Human Gene Functions Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 17805   Accepted: 9917 Description It is well known that a human gene can be considered as a sequence, consisting of four nucleotides, which are simply denoted by four…

Human Gene Functions Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 19573   Accepted: 10919 Description It is well known that a human gene can be considered as a sequence, consisting of four nucleotides, which are simply denoted by four…

Human Gene Functions Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 18007   Accepted: 10012 Description It is well known that a human gene can be considered as a sequence, consisting of four nucleotides, which are simply denoted by four…

大概作了一周,终于A了 类似于求最长公共子序列,稍有变形 当前序列 ch1 中字符为 a,序列 ch2 中字符为 b 则有 3 种配对方式: 1. a 与 b 2. a 与 - 3. - 与 b 动态转移方程: dp[i][j] = max(dp[i - 1][j - 1] + g(ch1[i],ch2[j]) , dp[i - 1][j] + g(ch1[i],‘-') , dp[i][j-1] + g('-',ch2[j])) 代码如下: #include<stdio.h> #includ…

最长公共子序列的变形 题目大意:给出两个基因序列,求这两个序列的最大相似度. 题目中的表格给出了两两脱氧核苷酸的相似度. 状态转移方程为: dp[i][j] = max(dp[i-1][j]+Similarity(s1[i], '-'),                     dp[i][j-1]+Similarity(s2[j], '-'),                     dp[i-1][j-1]+Similarity(s1[i], s2[j])); 注意边界的初始化. //#de…

题目大意: 两个字符串,可以再中间任何插入空格,然后让这两个串匹配,字符与字符之间的匹配有各自的分数,求最大分数 最长公共子序列模型. dp[i][j]表示当考虑吧串1的第i个字符和串2的第j个字符时,当前的最大分数,当前有3中可能, 1,i与j直接匹配,那么这个状态是由dp[i-1][j-1]转移过来的. 2,i与空格匹配,那么j就要与i-1匹配了,由dp[i-1][j]转移过来. 3,j与空格匹配,那么i就要与j-1匹配了,由dp[i][j-1]转移过来. dp的初始值,dp[i][0]与d…

Human Gene Functions Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 18053 Accepted: 10046 Description It is well known that a human gene can be considered as a sequence, consisting of four nucleotides, which are simply denoted by four let…

Human Gene Functions Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3103    Accepted Submission(s): 1761 Problem Description It is well known that a human gene can be considered as a sequence,…

POJ 1458 Common Subsequence(LCS最长公共子序列)解题报告 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87730#problem/F 题目: Common Subsequence Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 43388   Accepted: 17613 Description A subsequen…

Long Long Message Time Limit: 4000MS   Memory Limit: 131072K Total Submissions: 25752   Accepted: 10483 Case Time Limit: 1000MS Description The little cat is majoring in physics in the capital of Byterland. A piece of sad news comes to him these days…

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,…

POJ1458 Common Subsequence(最长公共子序列LCS) http://poj.org/problem?id=1458 题意: 给你两个字符串, 要你求出两个字符串的最长公共子序列长度. 分析: 本题不用输出子序列,非常easy,直接处理就可以. 首先令dp[i][j]==x表示A串的前i个字符和B串的前j个字符的最长公共子序列长度为x. 初始化: dp全为0. 状态转移: IfA[i]==B[j] then dp[i][j]= dp[i-1][j-1]+1 else dp[…

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…

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…