POJ 3080 Blue Jeans (求最长公共字符串) Description The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousands of contributors to map how the Earth was populated. As an IB…

题目:http://poj.org/problem?id=3080 水题,暴搜 #include <iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<stack> #include<queue> #include<iomanip> #include<cmath> #include<map> #include…

点击打开链接 Blue Jeans Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10243   Accepted: 4347 Description The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of…

Blue Jeans  Time Limit: 1000MS        Memory Limit: 65536K Total Submissions: 21078        Accepted: 9340 Description The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundred…

Blue Jeans [题目链接]Blue Jeans [题目类型]Java暴力 &题意: 就是求k个长度为60的字符串的最长连续公共子串,2<=k<=10 规定: 1. 最长公共串长度小于3不输出 2. 若出现等长的最长的子串,则输出字典序最小的串 &题解: 这个我刚开始用c++写的,真的是恶心到我了,啥都要自己实现,所以突然就想到用Java试试,结果还真的挺简单.思路就是在第1个串中找出所有子串,之后去下面找就行了. String.compareTo(Sting )是字符串…

Blue Jeans Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20966   Accepted: 9279 Description The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousa…

题目链接:http://poj.org/problem?id=3080 该题属于字符串处理中的串模式匹配问题.题目要求我们:给出一个DNA碱基序列,输出最长的相同的碱基子序列.(保证在所有的序列中都有出现) 这里采用了Brute Force算法(由于碱基序列的串长仅为60,规模比较小),这是模式匹配的一种最简单的做法. 设: 最长公共字串为ans,其长度为maxlen. m个碱基序列为p[0]...p[m-1].由于公共子序列是每个碱基序列的子串,因此不妨枚举p[0]的每一个可能的子串s.以s为…

题目:POJ3080 http://poj.org/problem?id=3080 题意:对于输入的文本串,输出最长的公共子串,如果长度相同,输出字典序最小的. 这题数据量很小,用暴力也是16ms,用后缀数组可以到0ms,但我不会XD. 暴力: #include<cstdio> #include<cstring> using namespace std; ][],ans[]; int main(){ int T,n; scanf("%d", &T); w…

[题目链接] http://poj.org/problem?id=3080 [题目大意] 求k个串的最长公共子串,如果存在多个则输出字典序最小,如果长度小于3则判断查找失败. [题解] 将所有字符串通过拼接符拼成一个串,做一遍后缀数组,二分答案,对于二分所得值,将h数组大于这个值的相邻元素分为一组,判断组内元素是否覆盖全字典,是则答案成立,对于答案扫描sa,输出第一个扫描到的子串即可. [代码] #include <cstdio> #include <cstring> #inclu…

题目 http://poj.org/problem?id=3080 题意 有m个(2<=m<=10)不包含空格的字符串,长度为60个字符,求所有字符串中都出现过的最长公共子序列,若该子序列长度小于3,输出"no significant commonalities",否则,输出字典序最小且长度最大的公共子序列. 思路 由于数据较小,其实可以使用比较暴力的思路,但这里为了复习后缀数组采用后缀数组的方法. 1. 将多个字符串合为一个目标字符串buff,为了防止程序无法分清字符串连…

题目:http://poj.org/problem?id=3080 Sample Input 3 2 GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 3 GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA GATACTAGATACTA…

题目网址:http://poj.org/problem?id=3080 思路: 以第一个DNA序列s为参考序列,开始做以下的操作. 1.将一个字母s[i]作为匹配串.(i为当前遍历到的下标) 2.遍历所有序列,看是否是所有序列的公共子串 3.是所有序列的子串的话,再往后增加一个字母,组成一个长度len+1的匹配串(设原先匹配串长度为len),重复步骤2 4.不是所有序列的子串的话,i=len+i;判断len是否大于3,是的话保存子串.len=0;重复步骤1.(为什么 i=len+i呢?因为len…

题意:给出m个字符串,找出其中的最长公共子序列,如果相同长度的有多个,输出按字母排序中的第一个. 思路:数据小,因此枚举第一个字符串的所有子字符串s,再一个个比较,是否为其它字符串的字串.判断是否为字串的时候,将s的字符依次与其他字符串的字符比较. #include <iostream> #include <stdio.h> #include <string.h> #include <stack> #include <algorithm> usi…

求出公共子序列  要求最长  字典序最小 枚举第一串的所有子串   然后对每一个串做KMP.找到目标子串 学会了   strncpy函数的使用   我已可入灵魂 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; char str[20][70]; char tmp[70],ans[70]; int f[70];…

Description The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousands of contributors to map how the Earth was populated. As an IBM researcher, you have been tas…

<题目链接> 题目大意: 就是求k个长度为60的字符串的最长连续公共子串,2<=k<=10 限制条件: 1.  最长公共串长度小于3输出   no significant commonalities 2.  若出现等长的最长的子串,则输出字典序最小的串 解题分析: 将第一个字串的所有子串枚举出来,然后用KMP快速判断该子串是否在所有主串中出现,如果都出现过,那么就按该子串的长度和字典序,不断更新答案,直到得到最终的最优解. #include <cstdio> #incl…

Time Limit: 1000MSMemory Limit: 65536K Total Submissions: 19026Accepted: 8466 Description The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousands of contributo…

题目大意: 求出这些DNA序列中的最长且字典序最小的公共子串. 思路分析: 二分长度的答案,去height中扫描这个长度是否满足,一旦满足就立即输出.这样就能够保证字典序最小了. #include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #define maxn 1005 using namespace std; char str[maxn]; int sa[m…

POJ 1488 题目大意:给定一篇文章,将它的左引号转成 ``(1的左边),右引号转成 ''(两个 ' ) 解题思路:水题,设置一个bool变量标记是左引号还是右引号即可 /* POJ 1488 Tex Quotes --- 水题 */ #include <cstdio> #include <cstring> int main() { #ifdef _LOCAL freopen("D:\\input.txt", "r", stdin); #…

题目:http://poj.org/problem?id=1035 还是暴搜 #include <iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<stack> #include<queue> #include<iomanip> #include<cmath> #include<map> #include&…

Description Word puzzles are usually simple and very entertaining for all ages. They are so entertaining that Pizza-Hut company started using table covers with word puzzles printed on them, possibly with the intent to minimise their client's percepti…

http://poj.org/problem?id=3461 直接KMP就好.水题 #include<cstdio> #include<cstring> const int MAXN=10000+10; const int MAXM=1000000+10; char P[MAXN],T[MAXM]; int f[MAXN],n,m,ans; void getFail() { f[0]=f[1]=0; for(int i=1;i<n;i++){ int j = f[i]; wh…

Biorhythms Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 110991   Accepted: 34541 Description Some people believe that there are three cycles in a person's life that start the day he or she is born. These three cycles are the physical,…

487-3279 Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 236746   Accepted: 41288 Description Businesses like to have memorable telephone numbers. One way to make a telephone number memorable is to have it spell a memorable word or phras…

Hangover Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 99450   Accepted: 48213 Description How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're as…

一.Description Suppose you are reading byte streams from any device, representing IP addresses. Your task is to convert a 32 characters long sequence of '1s' and '0s' (bits) to a dotted decimal format. A dotted decimal format for an IP address is form…

一.Description The king pays his loyal knight in gold coins. On the first day of his service, the knight receives one gold coin. On each of the next two days (the second and third days of service), the knight receives two gold coins. On each of the ne…

题意:给定一个n*m的矩阵,让你判断有多少个连通块. 析:用DFS搜一下即可. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring&…

题意:给你两个数,求所有的数位的积的和. 析:太水了,没的说,可以先输入边算,也可以最后再算,一样.. 代码如下: #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #inclu…

再思考一下好的方法,水过,数据太弱! 本来不想传的! #include <iostream> using namespace std; #define MAX 702 /*284K 422MS*/ typedef struct _point { int x; int y; }point; point p[MAX]; bool judge(point a,point b,point c) { return (a.y-b.y)*(c.x-b.x)-(c.y-b.y)*(a.x-b.x); } in…