poj 2955 Brackets 【区间dp 入门】

Brackets

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4789 Accepted: 2546

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 … a_n, 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, …, i_mwhere 1 ≤ i₁ < i₂ < … < i_m ≤ n, a_i1a_i2 … a_im 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.

Sample Input

((()))()()()([]]))[)(([][][)end

Sample Output

66406

10min1A好激动啊。不知道有没有bug。。。o(╯□╰)o 

题意：给定一个只由'['，']'，'('，')'组成的字符串，问你最大括号匹配的个数。

思路：设置dp[i][j]表示区间[i, j]里面最大括号匹配个数。

首先[i, j]由区间[i+1, j]推导而来，显然有dp[i][j] = dp[i+1][j]。

考虑加入字符str[i]对dp[i][j]的影响。

在from i+1 to j里面找，找到和字符str[i]匹配的字符下标k。

我们把区间[i+1, j]划分为三个子区间[i+1, k-1] [k, k] [k+1, j]，状态的转移就出来了。

dp[i][j] = max(dp[i][j], dp[i+1][k-1]+dp[k+1][j]+1)。

AC代码:

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <vector>#define INF 0x3f3f3f#define eps 1e-8#define MAXN (100+1)#define MAXM (100000)#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%.2lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while(a--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1using namespace std;char str[MAXN];int dp[MAXN][MAXN];bool judge(char a, char b){    return (a == '(' && b == ')') || (a == '[' && b == ']');}int main(){    while(Rs(str), strcmp(str, "end"))    {        int len = strlen(str);        CLR(dp, 0);        for(int i = len-1; i >= 0; i--)        {            for(int j = i; j < len; j++)            {                dp[i][j] = dp[i+1][j];                for(int k = i+1; k <= j; k++)                {                    if(judge(str[i], str[k]))                        dp[i][j] = max(dp[i][j], dp[i+1][k-1] + 1 + dp[k+1][j]);                }            }        }        Pi(dp[0][len-1]*2);    }    return 0;}