poj2955 (区间dp)

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 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 [([])].

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.

区间dp入门题，可我还是看题解才做出来。。。

用dp[i][j]表示区间i, j之间的最大匹配数。。。如果是s[i]与s[j]匹配，则dp[i][j] = dp[i + 1][j - 1] + 2，然后再将该区间分段算最大值（必须有，否则样例二过不了）

#include <vector>
#include <stdio.h>
#include <map>
#include <stdlib.h>
#include<ctype.h>
#include <iostream>
#include <queue>
#include <string.h>
#include <algorithm>
#define LL long long
using namespace std;
char s1[150];
int dp[150][150];
int main(int argc, char const *argv[]){
	while(~scanf("%s", s1)){
		int len = strlen(s1);
		if(strcmp(s1, "end") == 0){
			break;
		}

		
		memset(dp, 0, sizeof(dp));
		for(int i = 2; i <= len; i++){
			for(int j = 0; j < len; j++){
				int k = j + i - 1;
				if((s1[j] == '(' && s1[k] == ')') || (s1[j] == '[' && s1[k] == ']')){
					dp[j][k] = dp[j + 1][k - 1] + 2;
				}

				for(int x = j; x < k; x++){
					dp[j][k] = max(dp[j][k], dp[j][x] + dp[x + 1][k]);
				}
			}
		}

		printf("%d\n", dp[0][len - 1]);
	}
    return 0;
}