POJ 2955 Brackets 括号匹配 区间DP

题意：在一些括号中找到一个序列，里面的括号都两两配对。求序列最长长度。

区间DP，dp[i][j]为i~j的最大括号数，考虑第i个括号，有两种情况：不管i直接算dp[i + 1][j];找到和i匹配的右括号，分两边算并加起来，dp[i][j] = dp[i + 1][k - 1] + 2 + dp[k + 1][j]。

代码：

/**  Author:      illuz <iilluzen[at]gmail.com>*  Blog:        http://blog.csdn.net/hcbbt*  File:        poj2955.cpp*  Create Date: 2013-11-11 14:47:59*  Descripton:  intervel dp */#include <cstdio>#include <cstring>#define check(i, j) ((str[i] == '[' && str[j] == ']') || (str[i] == '(' && str[j] == ')'))#define max(a, b) ((a) > (b) ? (a) : (b))const int MAXN = 110;char str[MAXN];int dp[MAXN][MAXN], n;int solve(int i, int j) {if (dp[i][j]) return dp[i][j];if (j <= i) return 0;if (j == i + 1) {if (check(i, j))return dp[i][j] = 2;elsereturn 0;}dp[i][j] = solve(i + 1, j);for (int k = i + 1; k <= j; k++)if (check(i, k)) dp[i][j] = max(dp[i][j], solve(i + 1, k - 1) + solve(k + 1, j) + 2);return dp[i][j];}int main() {while (scanf("%s", str) && strcmp(str, "end")) {memset(dp, 0, sizeof(dp));printf("%d\n", solve(0, strlen(str) - 1));}return 0;}