题目大意 给定一个含有N个数的序列,要求你对一些数减掉或者加上某个值,使得序列变为非递减的,问你加减的值的总和最少是多少? 题解 一个很显然的结果就是,变化后的每一个值肯定是等于原来序列的某个值,因为只需要变为非递减的,所以对于某个数要么不变,要么变成左右附件的某个值.这样我们就可以根据前述条件得出DP方程了:dp[i][j]=min(dp[i][j-1],dp[i-1][j]+|a[i]-b[j]|)(a为原序列,b为排序后的序列),方程的意思是,把序列前i个数变为非递减序列并且以不超过b[j…

Description We give the following inductive definition of a “regular brackets” sequence: the empty sequence is a regular brackets sequence, if s is a regular brackets sequence, then (s) and [s] are regular brackets sequences, and if a and b are regul…

Brackets Sequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 35049   Accepted: 10139   Special Judge Description Let us define a regular brackets sequence in the following way: 1. Empty sequence is a regular sequence. 2. If S is a r…

题目链接:http://poj.org/problem?id=1141 题解:求已知子串最短的括号完备的全序列 代码: #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> using namespace std; #define ll long long ; const int INF=0x3f3f3f3f; ][]; ][]; ]; int Find(int x…

题意:给定一个括号序列,将它变成匹配的括号序列,可能多种答案任意输出一组即可.注意:输入可能是空串. 思路:D[i][j]表示区间[i, j]至少需要匹配的括号数,转移方程D[i][j] = min(D[i][k] + D[k+1][j], D[i][j]).     输入时,可能是空串应该用gets.fgets.getline,应注意换行符的吸收.每组数据前有一个换行符,输出的两组数据之间有换行. AC代码: #include<cstdio> #include<vector> #…

题目地址:Ural 1183 最终把这题给A了.. .拖拉了好长时间,.. 自己想还是想不出来,正好紫书上有这题. d[i][j]为输入序列从下标i到下标j最少须要加多少括号才干成为合法序列.0<=i<=j<len (len为输入序列的长度). c[i][j]为输入序列从下标i到下标j的断开位置.假设没有断开则为-1. 当i==j时.d[i][j]为1 当s[i]=='(' && s[j]==')' 或者 s[i]=='[' && s[j]==']'时,d…

http://blog.csdn.net/cc_again/article/details/10169643 http://blog.csdn.net/lijiecsu/article/details/7589877 如果有空串要用gets,scanf不能处理空串 #include <iostream> #include <string> #include <cstring> #include <cstdlib> #include <cstdio>…

题目链接 http://poj.org/problem?id=1141 Description Let us define a regular brackets sequence in the following way: 1. Empty sequence is a regular sequence. 2. If S is a regular sequence, then (S) and [S] are both regular sequences. 3. If A and B are reg…

Brackets Sequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 29520   Accepted: 8406   Special Judge Description Let us define a regular brackets sequence in the following way: 1. Empty sequence is a regular sequence. 2. If S is a re…

Description Let us define a regular brackets sequence in the following way: 1. Empty sequence is a regular sequence. 2. If S is a regular sequence, then (S) and [S] are both regular sequences. 3. If A and B are regular sequences, then AB is a regular…