poj 1080 dp如同LCS问题

题目链接:http://poj.org/problem?id=1080 #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> using namespace std; ; const int INF = 0x3f3f3f; int dp[maxn][maxn]; int A[maxn],B[maxn]; ][] = { {, , , , , }, {,,-,-,-,…

poj 1080 基因组（LCS）

Human Gene Functions Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 19376 Accepted: 10815 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…

poj 1080 dp

基因配对给出俩基因链和配对的值求配对值得最大值简单dp #include<iostream> #include<stdio.h> #include<string.h> using namespace std; ; ; char str1[maxa], str2[maxa]; ][] = { , -, -, -, -, -, , -, -, -, -, -, , -, -, -, -, -, , -, -, -, -, -,- }; int a1[maxa],…

poj 1080 zoj 1027(最长公共子序列变种)

http://poj.org/problem?id=1080 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=27 /* zoj 1027 poj 1080 思路: 三种状态,取最大值: s1[i]和s2[j]配 :dp[i-1][j-1]+cost[my[s1[i]]][my[s2[j]]]; s1[i]和'-' 配: dp[i-1][j]+cost[my[s1[i]]][my['-']]; s2[j]和'-' 配: dp[…

【POJ 1080】 Human Gene Functions

[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…

Poj 1936,3302 Subsequence(LCS)

一.Description(3302) Given a string s of length n, a subsequence of it, is defined as another string s' = su1su2...sum where 1 ≤ u1 < u2 < ... < um ≤ n and si is the ith character of s. Your task is to write a program that, given two strings s1 an…

POJ 1080：Human Gene Functions LCS经典DP

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…

POJ 1080（ LCS变形）

题目链接: http://poj.org/problem?id=1080 Human Gene Functions Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 20430 Accepted: 11396 Description It is well known that a human gene can be considered as a sequence, consisting of four nucleoti…

hdu 1513 && 1159 poj Palindrome (dp, 滚动数组， LCS)

题目以前做过的一道题, 今天又加了一种方法整理了一下..... 题意:给出一个字符串,问要将这个字符串变成回文串要添加最少几个字符. 方法一: 将该字符串与其反转求一次LCS,然后所求就是n减去最长公共子串的长度. 额,,这个思路还是不是很好想. LCS: #include<iostream> #include<cstring> #include<cstdio> using namespace std; +; char s1[maxn], s2[maxn]; ][…

dp poj 1080 Human Gene Functions

题目链接: http://poj.org/problem?id=1080 题目大意: 给两个由A.C.T.G四个字符组成的字符串,可以在两串中加入-,使得两串长度相等. 每两个字符匹配时都有个值,求怎样安排使得总的值最大,两个-不能匹配. 解题思路: 这题转化一下就是一个裸的最长公共子串问题,只不过要求匹配时长度一样. dp[i][j]表示第一串的第前i个字符和第二串的前j个字符匹配时,能达到的最大值. 初始化时注意dp[0][j]和dp[j][0]不能为零,为相应字符与-匹配时的总和. 代码:…

poj 1080 （LCS变形）

Human Gene Functions 题意: LCS: 设dp[i][j]为前i,j的最长公共序列长度: dp[i][j] = dp[i-1][j-1]+1;(a[i] == b[j]) dp[i][j] = max(dp[i][j-1],dp[i-1][j]); 边界:dp[0][j] = 0(j<b.size) ,dp[i][0] = 0(i< a.size); LCS变形: 设dp[i][j]为前i,j的最大价值: value(x, y)为比较价值: dp[i][j] = max(d…

poj 1159 (DP LCS)

滚动数组 + LCS // File Name: 1159.cpp // Author: Missa_Chen // Created Time: 2013年07月08日星期一 10时07分13秒 #include <iostream> #include <string> #include <algorithm> #include <cstdio> #include <cstring> #include <cmath> #includ…

poj 1080 Human Gene Functions(dp)

题目:http://poj.org/problem?id=1080 题意:比较两个基因序列,测定它们的相似度,将两个基因排成直线,如果需要的话插入空格,使基因的长度相等,然后根据那个表格计算出相似度. 题解: 考虑f[i][j]: ① s1取第i个,s2取第j个, f[i][j] = f[i-1][j-1]+value[m(s1[i])][m(s2[j])]; ② s1取第i个,s2用’-’, f[i][j] = f[i][j-1]+value[m(s1[i])][m(‘-’)];…

（线性dp，LCS） POJ 1458 Common Subsequence

Common Subsequence Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 65333 Accepted: 27331 Description A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..…

POJ - 1080 枚举 / DP

要求max{F/P},先枚举下界lowf,再贪心求符合约束条件的n个最小价值和记录F的离散值和去重可以大幅度常数优化 (本来想着用DP做的) (辣鸡POJ连auto都Complie Error) #include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #include<vector> #include<iterator> using name…

poj 1080 Human Gene Functions（lcs，较难）

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…

POJ 1080 Human Gene Functions 【dp】

题目大意:每次给出两个碱基序列(包含ATGC的两个字符串),其中每一个碱基与另一串中碱基如果配对或者与空串对应会有一个分数(可能为负),找出一种方式使得两个序列配对的分数最大思路:字符串动态规划的经典题,很容易想到状态dp[i][j],指第一个长度为i的串和第二个长度为j的串配对的最大分数.显然,这个状态可以由dp[i][j-1],dp[i-1][j],dp[i-1][j-1]三个子问题得到,即第一串最后一个字符对应空格.第二串最后一个字符对应空格和第一串第二串最后一个字符配对所得到的分数这三…

POJ 1080 Human Gene Functions -- 动态规划(最长公共子序列)

题目地址:http://poj.org/problem?id=1080 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 identifyi…

poj 1080

http://poj.org/problem?id=1080 知识点 :最长公共子序列要点: 转移方程 f[i][j] = max{ f[i-i][j]+score[s1[i-1]]['-'], f[i][j-1]+score['-'][s2[j-1]], f[i-1][j-1]+score[s1[i-1]][s2[j-1]]} #include <iostream> using namespace std; ][]; ][]; ],s2[]; void init(){ score[…

poj 1159 Palindrome 【LCS】

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…

DP：LCS（最长公共子串、最长公共子序列）

1. 两者区别约定:在本文中用 LCStr 表示最长公共子串(Longest Common Substring),LCSeq 表示最长公共子序列(Longest Common Subsequence). 子串要求在原字符串中是连续的,而子序列则没有要求.例如: 字符串 s1=abcde,s2=ade,则 LCStr=de,LCSeq=ade. 2. 求最长公共子串(LCStr) 算法描述:构建如下图的矩阵dp[][],当s1[i] == s2[j] 的时候,dp[i][j]=1:最后矩阵中斜对…

POJ 1159 Palindrome（LCS）

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…