POJ - 2955（2020.4.19训练C题）

Problem
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, andif a and b are regular brackets sequences, then ab is a regular brackets sequence.no other sequence is a regular brackets sequenceFor instance, all of the following character sequences are regular brackets sequences:(), [], (()), ()[], ()[()]while the following character sequences are not:(, ], )(, ([)], ([(]Given a brackets sequence of characters a1a2 … an, your goal is to find the length of the longest regular brackets sequence that is a subsequence of s. That is, you wish to find the largest m such that for indices i1, i2, …, im where 1 ≤ i1 < i2 < … < im ≤ n, ai1ai2 … aim is a regular brackets sequence.Given the initial sequence ([([]])], the longest regular brackets subsequence is [([])].

Input

The input test file will contain multiple test cases. Each input test case consists of a single line containing only the characters (, ), [, and ]; each input test will have length between 1 and 100, inclusive. The end-of-file is marked by a line containing the word “end” and should not be processed.

Output

For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.

题意：给定一个序列仅含’{‘或’}‘或’[‘或’]’,求其子串中的最长合法长度为多少

区间dp题，dp[i][j]代表下标i到j的合法最长长度，列举起点j和终点k，如果s[j]=s[k],则dp[j][k]=dp[j+1][k-1]+2,然后对段中分割，更新最大值

AC代码：

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int dp[105][105];
char s[105];
int main()
{
    int i, j, l, k;while (scanf("%s",s)!=EOF){
    if (strcmp(s, "end") == 0){
    return 0;}int len = strlen(s);//cout << len << endl;memset(dp, 0, sizeof(dp));for (i = 1; i < len; i++){
    for (j = 0; j + i < len; j++){
    k = i + j;if ((s[j]=='['&&s[k]==']')||(s[j]=='('&&s[k]==')')){
    //cout << "*" << endl;dp[j][k] = dp[j + 1][k - 1] + 2;}for (l = j; l < k; l++){
    dp[j][k] = max(dp[j][k], dp[j][l] + dp[l+1][k]);}}}cout << dp[0][len - 1] << endl;}
}