http://acm.hdu.edu.cn/showproblem.php?pid=5009 题目要求对空序列染成目标颜色序列,对一段序列染色的成本是不同颜色数的平方. 这题我们显然会首先想到用DP去解决,dp[i] = min( dp[i] , dp[j] + cost(i , j) ).但是枚举ij的复杂的高达n^2,不能接受,我们就考虑去优化它. 首先比较容易想到的是,去除连续重复元素,可以把它们当作同一个点处理. 此外在遍历j的过程中如果当前序列颜色不同的数量平方大于当前dp[i],显然…

Problem Description Lee has a string of n pearls. In the beginning, all the pearls have no color. He plans to color the pearls to make it more fascinating. He drew his ideal pattern of the string on a paper and asks for your help.  In each operation,…

Paint Pearls Problem Description   Lee has a string of n pearls. In the beginning, all the pearls have no color. He plans to color the pearls to make it more fascinating. He drew his ideal pattern of the string on a paper and asks for your help. In e…

转自:http://blog.csdn.net/accelerator_/article/details/39271751 吐血ac... 11668627 2014-09-16 22:15:24 Accepted 5009 1265MS 1980K 2290 B G++ czy   Paint Pearls Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Subm…

Paint Pearls Problem Description Lee has a string of n pearls. In the beginning, all the pearls have no color. He plans to color the pearls to make it more fascinating. He drew his ideal pattern of the string on a paper and asks for your help.  In ea…

首先把具有相同颜色的点缩成一个点,即数据离散化. 然后使用dp[i]表示涂满前i个点的最小代价.对于第i+1个点,有两种情况: 1)自己单独涂,即dp[i+1] = dp[i] + 1 2)从第k个节点之后(不包括k)到第i+1个节点一次涂完,且一起涂的节点共有num种颜色,即dp[i+1] = dp[k] + num * num 从而可以得到状态转移方程dp[i+1] = min(dp[i], dp[k] + num * num) 但是如果从后往前遍历每一个k,会超时. 因此我们可以使用双向链…

题目链接:https://cn.vjudge.net/problem/HDU-5009 题意 给一串序列,可以任意分割多次序列,每次分割的代价是被分割区间中的数字种数. 求分割区间的最小代价.n<=5e4 例:1 3 3 答:2 思路 今天刚讲的dp题目,这次讲课收益非常,写个题解. 状态和转移是dp[i]=min(dp[j]+cnt[j-1][i]^2) 会发现这样绝对超时,又发现这里的j可以贪心的跳着选择(因为种数相同时,左端点应越小越好). 这里的跳着选择想了半天没什么思路,又是最后看了题…

Dicripntion Lee has a string of n pearls. In the beginning, all the pearls have no color. He plans to color the pearls to make it more fascinating. He drew his ideal pattern of the string on a paper and asks for your help. In each operation, he selec…

Paint Pearls 思路: 离散化+dp+剪枝: dp是个n方的做法: 重要就在剪枝: 如果一个长度为n的区间,有大于根号n种颜色,还不如一个一个涂: 来,上代码: #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define maxn 50005 int n,ai[maxn],dp[maxn],ls[…

http://acm.hdu.edu.cn/showproblem.php?pid=5009 2014网络赛 西安 比较难的题 Paint Pearls Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1951    Accepted Submission(s): 631 Problem Description Lee has a str…