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1. 最长上升子序列 [题目描述] 给定N个数,求这N个数的最长上升子序列的长度. [样例输入] 7 2 5 3 4 1 7 6 [样例输出] 4 第一种解法:时间复杂度O(n^2), 状态设计:DP[ i ]代表以A[ i ]结尾的LIS的长度 状态转移:DP[ i ]=max{ DP[ j ]+1 ,DP[ i ] }(1<=j< i,A[j]< A[i]). 基本上思想,每次碰到一个数,就从头遍历,看插在哪个后面的值最大就保留那一个. #include <iostream&g…
https://www.luogu.org/problemnew/show/P1004 这道题分类到简单dp但是感觉一点都不简单……这种做两次的dp真的不是很懂怎么写.假如是贪心做两次,感觉又不能证明这是最优的. 直接看题解,题解要设置4个维度,两个人同时走……但是怎么避免同一个物品被两个人拿呢? 设置dp[i][j][k][l]表示第1个人走到ij,第2个人走到kl能实现的最大的取法,再加一维[p]表示现在已经取了00,01,02,10,11,12,20,21,22个物品,然后dp的内容是一个…
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…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2571 简单dp, dp[n][m] +=(  dp[n-1][m],dp[n][m-1],d[i][k] ) k 为m的因子 PS:0边界要初始为负数(例如-123456789)越大越好 代码: #include <stdio.h> #include <string.h> int dp[25][1005]; #define max(x,y) x > y ? x : y int m…