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 e…

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…

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,…

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…

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…

Paint Pearls 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5009 dp+双向链表优化 看到题目,很自然地可以定义状态:dp[i]表示涂好a[0...i]的字符串,花费的最小代价. 状态转移方程:dp[i]=min(dp[i],dp[j]+num2),其中num=从a[j]到a[i]不同的数字个数. 时间复杂度为O(n2),对于n=50000的数据,明显会T. 于是,我们需要进行优化.注意到状态数无法化简,考虑优化转移复杂度. 当区间…

Paint Pearls Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 3239    Accepted Submission(s): 1052 Problem Description Lee has a string of n pearls. In the beginning, all the pearls have no color…