题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1087 题目大意: 求递增子序列最大和 思路: 直接dp就可以求解,dp[i]表示以第i位结尾的递增子序列最大和,初始化dp[i] = a[i],转移方程dp[i] = max(dp[i], a[i] + dp[j])如果j < i && a[j] < a[i] #include<cstdio> #include<cstring> #include<i…

[问题描述] 给定长度为n的正整数序列a1,a2,…,an. 求一个递增的子序列,和最大. [输入] 第一行,n,表示给定序列的个数. 第二行,n个用空格隔开的正整数. [输出] 递增子序列的最大和. [数据范围限制] n<=1000,0<ai<=109. ..] of longint; f:..] of longint; n,i,j,ans:longint; function max(x,y:longint):longint; begin if x>y then exit(x)…

Super Jumping! Jumping! Jumping! Problem Description Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now. The game ca…

Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now. The game can be played by two or more than two players. It consi…

Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 32564    Accepted Submission(s): 14692 Problem Description Nowadays, a kind of chess game called “Super Jumping!…

layout: post title: 「kuangbin带你飞」专题十二 基础DP author: "luowentaoaa" catalog: true tags: mathjax: true - kuangbin - 动态规划 传送门 A.HDU1024 Max Sum Plus Plus 题意 给你N个数,然后你分成M个不重叠部分,并且这M个不重叠部分的和最大. 思路 动态规划最大m字段和,dp数组,dp[i][j]表示以a[j]结尾的,i个字段的最大和 两种情况:1.第a[j…

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!…

dp HDU - 1257 最少拦截系统 最长递增子序列 #include<iostream> using namespace std; const int maxn=1e7; int a[maxn],dp[maxn],n; int main() { while(cin>>n) { ; i <= n; i++)cin >> a[i]; ; i <= n; i++)dp[i] = ; ; i <= n; i++) ; j < i; j++) { i…

HDU 1024第一遍水过,没有体会到这个题的奥妙,思考了很久终于体会了.大概意思是求把序列分成m段的子序列,并不一定要覆盖完,求子序列和的最大值我们首先要写出基本的动态转移方程: DP:dp[ i ] [ j ] =max ( dp[ i - 1 ] [ 1~j-1 ]+a[ j ],dp[ i ] [ j - 1 ]+a[ j ] ) 其中dp[ i ][ j ]其中i,j表示,第i个集合,取前j 个数. 意思是,dp[ i ][ j ]是由两个状态转移过来的,一个是把新的数归入i集合,还有…

队友的建议,让我去学一学kuangbin的基础dp,在这里小小的整理总结一下吧. 首先我感觉自己还远远不够称为一个dp选手,一是这些题目还远不够,二是定义状态的经验不足.不过这些题目让我在一定程度上加深了对dp的理解,但要想搞好dp,还需要多多练习啊. HDU - 1024 开场高能 给出一个数列,分成m段,求这m段的和的最大值,dp[i][j]表示遍历到第i个数,已经划分了j段,对于每一个数有两种决策,加入上一个区间段,自己成为一个区间段,故dp[i][j] = max(dp[i-1][j]+…