POJ 2955 Brackets（括号匹配-区间DP）

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题目链接：http://poj.org/problem?id=2955 Brackets Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8302   Accepted: 4417 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 ofs. That is, you wish to find the largestm 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 6 6 4 0 6 Source Stanford Local 2004 题意：求最大括号匹配数，( )、[ ]. 若i、j不匹配，则dp[i][j] = dp[i + 1][j]；若i、j匹配，则dp[i][j] = max{dp[i][k] + dp[k + 1][j]} + 2. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; char s[110]; int dp[110][110]; int main() { while(~scanf("%s", s + 1) && s[1] != 'e') { int l = strlen(s + 1); memset(dp, 0, sizeof(dp)); for(int i = l - 1; i >= 1; i --) for(int j = i + 1; j <= l; j ++) { dp[i][j] = dp[i + 1][j]; for(int k = i + 1; k <= j; k ++) { if(s[i] == '(' && s[k] == ')' || s[i] == '[' && s[k] == ']') dp[i][j] = max( dp[i][j], dp[i + 1][k - 1] + dp[k + 1][j] + 2); } } printf("%d\n", dp[1][l]); } return 0; }