http://poj.org/problem?id=1699 题意:给你n个长度为L的序列,求包含这几个序列的最短长度. 先预处理每两个序列之间的关系,然后dfs枚举就行. #include <cstdio> #include <cstring> #include <algorithm> #define maxn 500 using namespace std; <<; char str[maxn][maxn]; int c[maxn][maxn]; int

id=1699" target="_blank" style="">题目链接:poj 1699 Best Sequence 题目大意:给定N个DNA序列,问说最少多长的字符串包括全部序列. 解题思路:AC自己主动机+状压DP,先对字符串构造AC自己主动机.然后在dp[s][i]表示匹配了s.移动到节点i时候的最短步数. #include <cstdio> #include <cstring> #include <queu

题目: http://poj.org/problem?id=1699 无意间A了..超时一次,加了一句 if(len > ans)return; 然后就A了,dfs题,没有太多好说的,代码写的效率高一点就行. #include <stdio.h> #include <string.h> ][]; ]; ]; int ans, n; void dfs(int cnt, int x, char tmp[], int len) { if(len > ans)return; v

意甲冠军:看图片是晶莹剔透的,正确的, N连接到第一序列(同样的序列部分).总序列获得最短. 主题链接:http://poj.org/problem?id=1699 ~~~~ 思路就是:将N个序列首尾相连能重合的长度求粗来. 然后DFS枚举每种首尾相连的情况. #include<cstdio> #include<cstring> #include<algorithm> #define N 22 #define INF 0x7fffffff using namespace

題目鏈接 題意 : 將幾個片段如圖所示方法縮成一個序列,求出最短這個序列. 思路 : 其實我也不知道怎麼做.....看網上都用了DP.....但是我不會.....這個DP不錯,還有用KMP+状压DP做的 #include <iostream> #include <stdio.h> #include <string.h> #include <string> using namespace std; ] ; ][],ans ,n,vis[]; void f(in

题目链接 题意:给出几个基因片段,要求你将它们排列成一个最短的序列,序列中使用了所有的基因片段,而且不能翻转基因. 分析:先计算出add数组,再dfs枚举. 空间复杂度O(n*n),  最坏时间复杂度 O(n^n),但是剪枝以后很快,因为好多搜不到后面,搜不到第n层. #include <iostream> #include <cstring> #include <cstdlib> #include <cmath> #include <cstdio&g

[poj P1141] Brackets Sequence Time Limit: 1000MS   Memory Limit: 65536K   Special Judge Description Let us define a regular brackets sequence in the following way: 1. Empty sequence is a regular sequence. 2. If S is a regular sequence, then (S) and [

一.题目大意 有这样一个序列包含S1,S2,S3...SK,每一个Si包括整数1到 i.求在这个序列中给定的整数n为下标的数. 例如,前80位为11212312341234512345612345671234567812345678912345678910123456789101112345678910,第8位为2. 二.题解 动手做这道题之前要了解所求的是第n位而不是某一个Si中的第几个数,一位数占一位,两位数占两位,三位占三位...由于每一组都比前一组多一个数,如果这个数是一位数,则该组数的

 名字是法雷数列其实是欧拉phi函数              Farey Sequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11246   Accepted: 4363 Description The Farey Sequence Fn for any integer n with n >= 2 is the set of irreducible rational numbers a/b with 0 <

A - Farey Sequence Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2478 Description The Farey Sequence Fn for any integer n with n >= 2 is the set of irreducible rational numbers a/b with 0 &l

DNA Sequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9899   Accepted: 3717 Description It's well known that DNA Sequence is a sequence only contains A, C, T and G, and it's very useful to analyze a segment of DNA Sequence,For exa

Pots Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9963   Accepted: 4179   Special Judge Description You are given two pots, having the volume of A and B liters respectively. The following operations can be performed: FILL(i)        fi

Max Sequence Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 17678   Accepted: 7401 Description Give you N integers a1, a2 ... aN (|ai| <=1000, 1 <= i <= N).  You should output S.  Input The input will consist of several test cases.

DNA Sequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9889   Accepted: 3712 Description It's well known that DNA Sequence is a sequence only contains A, C, T and G, and it's very useful to analyze a segment of DNA Sequence,For exa

POJ - 2676 : http://poj.org/problem?id=2676: 解题思想 (大力出奇迹): 1. 依次在空格里面填上“1~9”,并检查这个数字是否合法(其所在的行.列,以及3X3的子区域里不存在重复的数字).如果合法,则前进到下一个格子. 2. 如果在某个格子里,从“1”到“9”都不合法,这说明前面某个格子填错了.这时就回退到上一格,继续试.例如,如果上一格已填的数字是3,就继续试4,5,6,… 是否合法.如果找到一个合法的数字,则又前进到下一格.如果找不到,说明前面还

Brackets Sequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 29520   Accepted: 8406   Special Judge Description Let us define a regular brackets sequence in the following way: 1. Empty sequence is a regular sequence. 2. If S is a re

Farey Sequence Time Limit: 1000MS   Memory Limit: 65536K       Description The Farey Sequence Fn for any integer n with n >= 2 is the set of irreducible rational numbers a/b with 0 < a < b <= n and gcd(a,b) = 1 arranged in increasing order. Th

题目地址:http://poj.org/problem?id=1019 Number Sequence Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 35680   Accepted: 10287 Description A single positive integer i is given. Write a program to find the digit located in the position i in

d(A) = max{sum(a[s1]..a[t1]) + sum(a[s2]..a[t2]) | 1<=s1<=t1<s2<=t2<=n} 即求两个子序列和的和的最大值. 为单个区间子序列和的最大值的变形. 左边的从左向右扫描,而右边的从右向左扫描即可. #include<cstdio> #include<algorithm> using namespace std; const int MAXN=50000+10; int left[MAXN],

题目链接:http://poj.org/problem?id=1141 题解:求已知子串最短的括号完备的全序列 代码: #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> using namespace std; #define ll long long ; const int INF=0x3f3f3f3f; ][]; ][]; ]; int Find(int x