POJ2955 Brackets 【区间DP】

题目链接：POJ-2955

kuangbin带你飞【专题二十二】 区间DP

题目描述：

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 regular brackets sequences, then ab is a regular brackets sequence.
no other sequence is a regular brackets sequence
For 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 [([])].

思路:

对于这道题，我们可以定义dp[i][j],表示从i到j的最大括号匹配数；
然后我们可以发现，dp[i][j] = dp[i][k] + dp[k+1][j]
我们每次更新前需要判断 s[i]是否等于s[j],如果等于 则将dp[i][j]赋值为dp[i+1][j-1]+1,不等于则赋值为dp[i+1][j-1]，然后枚举断点更新dp[i][j]即可;

代码:

#include <iostream>
using namespace std;#define INF 0x3f3f3f3f
#define long long long
#define MAX_N 110int dp[MAX_N][MAX_N];
string s;bool judge(int x, int y)
{
    if( (s[x] == '(' && s[y] == ')') || (s[x] == '[' && s[y] == ']') )return true;return false;
}int main()
{
    while(cin >> s, s != "end"){
    s = ' ' + s;for(int k = 2; k <= s.length(); k++){
    for(int i = 1; i + k - 1 <= s.length(); i++){
    int ends = i + k - 1;dp[i][ends] = judge(i, ends) ? dp[i + 1][ends - 1] + 2 : dp[i + 1][ends - 1];for(int j = i; j < ends; j++){
    dp[i][ends] = max(dp[i][ends], dp[i][j] + dp[j + 1][ends]);}}}cout << dp[1][s.length()] << endl;}return 0;
}