POJ 2955 Brackets (區間dp 括弧匹配)，poj2955

POJ 2955 Brackets (區間dp 括弧匹配)，poj2955

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3951		Accepted: 2078

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, thenab 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 ofs. That is, you wish to find the largest m such that for indicesi1, 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.

Sample Input

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

Sample Output

66406

Source

Stanford Local 2004

題目連結：http://poj.org/problem?id=2955

題目大意：給一個括弧序列，問序列中合法的括弧最多有多少個，若A合法，則[A]，(A)均合法，若A,B合法則AB也合法

題目分析：和POJ 1141那道經典括弧匹配類似，這題更簡單一些，想辦法把問題轉化，既然要求最大的括弧匹配數，我們考慮加最少的括弧，使得整個序列合法，這樣就轉變成1141那題，開下腦動類比二分圖最大匹配的性質，最大匹配＋最大獨立集＝點數，顯然要加入最少的點使序列合法，則加的最少的點數即為|最大獨立集|，我們要求的是原序列的|最大匹配|，以上純屬yy，下面給出轉移方程，和1141一模一樣
dp[i][i] = 1;
然後枚舉區間長度
1）外圍匹配：dp[i][j] = dp[i + 1][j - 1];
2）外圍不匹配，枚舉分割點：dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j]); (i <= k < j)

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int const INF = 0x3fffffff;char s[205];int dp[205][205];int main(){    while(scanf("%s", s) != EOF && strcmp(s, "end") != 0)    {        int len = strlen(s);        memset(dp, 0, sizeof(dp));        for(int i = 0; i < len; i++)            dp[i][i] = 1;        for(int l = 1; l < len; l++)        {            for(int i = 0; i < len - l; i++)            {                int j = i + l;                dp[i][j] = INF;                if((s[i] == '(' && s[j] == ')') || (s[i] == '[' && s[j] == ']'))                    dp[i][j] = dp[i + 1][j - 1];                for(int k = i; k < j; k++)                    dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j]);            }        }        printf("%d\n", len - dp[0][len - 1]);    }}

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