[NYIST15]括号匹配（二）（区间dp）

题目链接：http://acm.nyist.edu.cn/JudgeOnline/problem.php?pid=15

经典区间dp，首先枚举区间的大小和该区间的左边界，这时右边界也可计算出来。首先初始化一个匹配，那就是看看这两个括号是否匹配，即：

(s[i] == '(' && s[j] == ')') || (s[i] == '[' && s[j] == ']') ? dp(i,j) = dp(i+1,j-1)+2) : dp(i,j) = 0

接下来枚举i和j中间的所有点，更新dp(i,j)=max(dp(i,j), dp(i+m)+dp(m+1,j))寻找可能更优的匹配。

 /*
 ━━━━━┒ギリギリ♂ eye！
 ┓┏┓┏┓┃キリキリ♂ mind！
 ┛┗┛┗┛┃＼○／
 ┓┏┓┏┓┃ /
 ┛┗┛┗┛┃ノ)
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┃┃┃┃┃┃
 ┻┻┻┻┻┻
 */
 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>
 using namespace std;
 #define fr first
 #define sc second
 #define cl clear
 #define BUG puts("here!!!")
 #define W(a) while(a--)
 #define pb(a) push_back(a)
 #define Rint(a) scanf("%d", &a)
 #define Rll(a) scanf("%I64d", &a)
 #define Rs(a) scanf("%s", a)
 #define Cin(a) cin >> a
 #define FRead() freopen("in", "r", stdin)
 #define FWrite() freopen("out", "w", stdout)
 #define Rep(i, len) for(int i = 0; i < (len); i++)
 #define For(i, a, len) for(int i = (a); i < (len); i++)
 #define Cls(a) memset((a), 0, sizeof(a))
 #define Clr(a, x) memset((a), (x), sizeof(a))
 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
 #define lrt rt << 1
 #define rrt rt << 1 | 1
 #define pi 3.14159265359
 #define RT return
 #define lowbit(x) x & (-x)
 #define onenum(x) __builtin_popcount(x)
 typedef long long LL;
 typedef long double LD;
 typedef unsigned long long ULL;
 typedef pair<int, int> pii;
 typedef pair<string, int> psi;
 typedef pair<LL, LL> pll;
 typedef map<string, int> msi;
 typedef vector<int> vi;
 typedef vector<LL> vl;
 typedef vector<vl> vvl;
 typedef vector<bool> vb;

 const int maxn = ;
 int dp[maxn][maxn];
 char s[maxn];
 int n;

 int main() {
     // FRead();
     int T;
     Rint(T);
     W(T) {
         Rs(s); n = strlen(s);
         Cls(dp);
         For(k, , n+) {
             Rep(i, n-k+) {
                 dp[i][i+k-] = ;
                 int j = i + k - ;
                 if((s[i] == '(' && s[j] == ')') || (s[i] == '[' && s[j] == ']')) {
                     dp[i][j] = dp[i+][j-] + ;
                 }
                 For(m, i, j) {
                     dp[i][j] = max(dp[i][j], dp[i][m] + dp[m+][j]);
                 }
             }
         }
         printf("%d\n", n - dp[][n-]);
     }
     RT ;
 }