Mix and Build(简单DP)】的更多相关文章

Mix and Build Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 3936 Accepted: 1203 Case Time Limit: 2000MS Special Judge Description In this problem, you are given a list of words (sequence of lower case letters). From this list, find the l…
学习Trie树中,所以上网搜一下Trie树的题,找到这个,人家写着是简单dp,那我就想着能学习到什么Trie树上的dp,但最后发现根本好像跟Trie树没有什么联系嘛... 题意就是给你很多个字符串(长度<20),然后如果两个字符串在排序完后,左边的一个+一个字符能变成右边的,这两个字符串连上一条边,然后求最长的边.我想了半天怎么跟Trie不搭边... 一个自然的想法是这样的,记d[i]为序号为i的字符串作为结束字符串的最长长度,我先把所有字符串根据长度由小到大排序,然后对每个字符串k,我每次把起…
题意: 就是现在给出m个串,每个串都有一个权值,现在你要找到一个长度不超过n的字符串, 其中之前的m个串每出现一次就算一次那个字符串的权值, 求能找到的最大权值的字符串,如果存在多个解,输出最短的字典序最小的串. 当最大全权值为0时输出空串. 输入最多100个子串,权值为不超过100的正整数. 每个子串长度至少为1,不超过10, n <= 50 如果不考虑方案输出,这题就变得相当简单了. dp[i][j]表示走到长度为 i 的时候 ,到AC自动机 j 这个节点所获得的最大权值和. 我一开始的做法…
Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 33384    Accepted Submission(s): 15093 Problem Description Nowadays, a kind of chess game called “Super Jumping!…
题目链接:http://codeforces.com/problemset/problem/455/A 给你n个数,要是其中取一个大小为x的数,那x+1和x-1都不能取了,问你最后取完最大的和是多少. 简单dp,dp[i]表示取i时zui最大和为多少,方程为dp[i] = max(dp[i - 1] , dp[i - 2] + cont[i]*i). #include <bits/stdc++.h> using namespace std; typedef __int64 LL; ; LL a…
Problem H. ICPC QuestTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100500/attachments Description Noura Boubou is a Syrian volunteer at ACM ACPC (Arab Collegiate Programming Contest) since 2011. She graduated from Tishreen Un…
题目链接 这道题也是简单dp里面的一种经典类型,递推式就是dp[i] = min(dp[i-150], dp[i-200], dp[i-350]) 代码如下: #include<iostream> #include <stdio.h> using namespace std; ]; int main() { ; i < ; i++) dp[i] = i; ; i < ; i++) { int minn; ) dp[i] = dp[i - ]; ) dp[i] = min…
J - 简单dp Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Description It is a little known fact that cows love apples. Farmer John has two apple trees (which are conveni…
I - 简单dp 例题扩展 Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HD…
题意:给你n种花,m个盆,花盆是有顺序的,每种花只能插一个花盘i,下一种花的只能插i<j的花盘,现在给出价值,求最大价值 简单dp #include <iostream> #include<cstdio> #include<cstring> using namespace std; #define N 110 int dp[N][N],a[N][N]; int main(int argc, char** argv) { int n,m,i,j; while(sca…