HDU - 2476 思路:分解问题,先考虑从一个空串染色成 B串的最小花费 ,区间DP可以解决这个问题 具体的就是,当 str [ l ] = = str [ r ]时 dp [ L ] [ R ] = min (dp [ L + 1] [ R ],dp [ L ] [ R-1 ] ) 其他情况可以选择任意一个断点 tmp = min ( tmp , dfs ( l  ,k ) + dfs ( k+1 , r ) ); 接下来就是第二步 现在A串不是空串 ,但然把它当做空串去染是没有问题的,…

String painter Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2068    Accepted Submission(s): 908 Problem Description There are two strings A and B with equal length. Both strings are made up o…

题意: 给出两个串s1和s2,一次只能将一个区间刷一次,问最少几次能让s1=s2 例如zzzzzfzzzzz,长度为11,我们就将下标看做0~10 先将0~10刷一次,变成aaaaaaaaaaa 1~9刷一次,abbbbbbbbba 2~8:abcccccccba 3~7:abcdddddcba 4~6:abcdeeedcab 5:abcdefedcab 这样就6次,变成了s2串了 其 实如果a串是空串的话,我们可以写出这样的区间dp方程:设dp[i][j]表示从i到j至少要变多少次,则有dp[…

题解 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; + ; int n; char s1[maxn], s2[maxn]; int d[maxn][maxn], a[maxn]; int main() { , s2 + ) == ) { n = strlen(s1 + ); ; i <= n; i++)…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=2476 题目大意: 给出两个等长的串S, T, 要将S变成T, 每次可以把S的连续的一段变成相同的字母,求最少操作数. 这题网上看了好多题解,理解了好久, 记录一下我的理解吧. 首先求出把空串变成T的最少次数. dp[i][j] 表示把空串变成T[i ... j]的最少次数. 首先dp[i][j] = dp[i + 1][j]. 然后有一个性质.如果两次染色的区间有交, 那么小的区间一定完全包含于大…

HDU 2829 区间DP & 前缀和优化 & 四边形不等式优化 n个节点n-1条线性边,炸掉M条边也就是分为m+1个区间 问你各个区间的总策略值最少的炸法 就题目本身而言,中规中矩的区间DP问题 d p[i][j]表示前i个节点,分为j个区间的最优策略值 cost[i][j]为从i到j节点的策略值 所以dp[i][j] = min(dp[k-1][j-1] + cost[k][i] 但是复杂度太高了 可以优化的地方有: cost数组值得求取: 考虑到cost(i,j)=ΣAxAy (i≤…

You Are the One Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3348    Accepted Submission(s): 1524 Problem Description The TV shows such as You Are the One has been very popular. In order to m…

题目链接 http://acm.split.hdu.edu.cn/showproblem.php?pid=4293 Problem Description After the regional contest, all the ACMers are walking alone a very long avenue to the dining hall in groups. Groups can vary in size for kinds of reasons, which means, sev…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4632 注意到任意一个回文子序列收尾两个字符一定是相同的,于是可以区间dp,用dp[i][j]表示原字符串中[i,j]位置中出现的回文子序列的个数,有递推关系: dp[i][j]=dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1] 如果i和j位置出现的字符相同,那么dp[i][j]可以由dp[i+1][j-1]中的子序列加上这两个字符构成回文子序列,也就是 dp[i][j]+=dp…

Palindrome subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65535 K (Java/Others)Total Submission(s): 2858    Accepted Submission(s): 1168 Problem Description In mathematics, a subsequence is a sequence that can be derived f…