Brackets Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5424   Accepted: 2909 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…

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…

很好的区间DP题. 需要注意第一种情况不管是否匹配,都要枚举k来更新答案,比如: "()()()":dp[0][5]=dp[1][4]+2=4,枚举k,k=1时,dp[0][1]+dp[2][5]=6,最后取最大值6. 第一层d相当于"长度"的含义,第二层枚举i,j就可以用i+d表示,通过这种方式枚举区间左右端点. 1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring>…

给一个括号序列,求有几个括号是匹配的. dp[i][j]表示序列[i,j]的匹配数 dp[i][j]=dp[i+1][j-1]+2(括号i和括号j匹配) dp[i][j]=max(dp[i][k]+dp[k+1][j])(i<=k<j) #include<cstdio> #include<cstring> #include<algorithm> using namespace std; ]; ][]; int main(){ ]!='e'){ int n=s…

题意:给定一个字符串 输出回文子序列的个数    一个字符也算一个回文 很明显的区间dp  就是要往区间小的压缩! #include<bits/stdc++.h> using namespace std; //input #define rep(i,x,y) for(int i=(x);i<=(y);++i) #define RI(n) scanf("%d",&(n)) #define RII(n,m) scanf("%d%d",&…

题目链接:https://vjudge.net/problem/POJ-2955 Brackets Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9630   Accepted: 5131 Description We give the following inductive definition of a “regular brackets” sequence: the empty sequence is a regu…

题目链接:http://poj.org/problem?id=2955 题意:给定字符串 求括号匹配最多时的子串长度. 区间dp,状态转移方程: dp[i][j]=max ( dp[i][j] , 2+dp[i+1][k-1]+dp[k+1][j] ); 代码: #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> using namespace std; #def…

//Accepted 200 KB 63 ms //区间dp //dp[i][j] 从i位到j位能得到的最大匹配数 //dp[i][j]=max(dp[i+1][j-1] (s[i-1]==s[j-1]),dp[i][k]+dp[k+1][j])i<=k<j #include <cstdio> #include <cstring> #include <iostream> using namespace std; ; int dp[imax_n][imax_n…

题目大意: 给一个由,(,),[,]组成的字符串,其中(),[]可以匹配,求最大匹配数 题解:区间dp: dp[i][j]表示区间 [i,j]中的最大匹配数 初始状态 dp[i][i+1]=(i,i+1可以匹配)?2:0 状态转移见代码 代码: #include <iostream> #include <stdio.h> #include<string.h> #include<algorithm> #include<string> #includ…

题目链接[http://poj.org/problem?id=2955] 题意:[].()的匹配问题,问一个[]()串中匹配的字符数,匹配方式为[X],(X),X为一个串,问一个长度为N(N<=100)串中最多的匹配字符个数. 思路:区间DP,dp[l][r]的意思是区间[l,r]的最大匹配数,预处理长度为2的所有区间的最大匹配数,然后由长度为2的区间推出长度为3的所有区间的最大匹配数,由长度为3的区间......在处理区间[l,r]的时候,如果s[l]与s[r]相匹配,那么dp[l][r]=d…