[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。

/*
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┓┏┓┏┓┃ /
┛┗┛┗┛┃ノ)
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┻┻┻┻┻┻
*/
#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 = 1100;
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, 2, n+1) {
Rep(i, n-k+1) {
dp[i][i+k-1] = 0;
int j = i + k - 1;
if((s[i] == '(' && s[j] == ')') || (s[i] == '[' && s[j] == ']')) {
dp[i][j] = dp[i+1][j-1] + 2;
}
For(m, i, j) {
dp[i][j] = max(dp[i][j], dp[i][m] + dp[m+1][j]);
}
}
}
printf("%d\n", n - dp[0][n-1]);
}
RT 0;
}