POJ-3080-Blue jeans(KMP, 暴力)】的更多相关文章

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…
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…
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…
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…
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…
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…
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…
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];…
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…