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。
1 /* 2 ━━━━━┒ギリギリ♂ eye! 3 ┓┏┓┏┓┃キリキリ♂ mind! 4 ┛┗┛┗┛┃\○/ 5 ┓┏┓┏┓┃ / 6 ┛┗┛┗┛┃ノ) 7 ┓┏┓┏┓┃ 8 ┛┗┛┗┛┃ 9 ┓┏┓┏┓┃ 10 ┛┗┛┗┛┃ 11 ┓┏┓┏┓┃ 12 ┛┗┛┗┛┃ 13 ┓┏┓┏┓┃ 14 ┃┃┃┃┃┃ 15 ┻┻┻┻┻┻ 16 */ 17 #include <algorithm> 18 #include <iostream> 19 #include <iomanip> 20 #include <cstring> 21 #include <climits> 22 #include <complex> 23 #include <cassert> 24 #include <cstdio> 25 #include <bitset> 26 #include <vector> 27 #include <deque> 28 #include <queue> 29 #include <stack> 30 #include <ctime> 31 #include <set> 32 #include <map> 33 #include <cmath> 34 using namespace std; 35 #define fr first 36 #define sc second 37 #define cl clear 38 #define BUG puts("here!!!") 39 #define W(a) while(a--) 40 #define pb(a) push_back(a) 41 #define Rint(a) scanf("%d", &a) 42 #define Rll(a) scanf("%I64d", &a) 43 #define Rs(a) scanf("%s", a) 44 #define Cin(a) cin >> a 45 #define FRead() freopen("in", "r", stdin) 46 #define FWrite() freopen("out", "w", stdout) 47 #define Rep(i, len) for(int i = 0; i < (len); i++) 48 #define For(i, a, len) for(int i = (a); i < (len); i++) 49 #define Cls(a) memset((a), 0, sizeof(a)) 50 #define Clr(a, x) memset((a), (x), sizeof(a)) 51 #define Full(a) memset((a), 0x7f7f7f, sizeof(a)) 52 #define lrt rt << 1 53 #define rrt rt << 1 | 1 54 #define pi 3.14159265359 55 #define RT return 56 #define lowbit(x) x & (-x) 57 #define onenum(x) __builtin_popcount(x) 58 typedef long long LL; 59 typedef long double LD; 60 typedef unsigned long long ULL; 61 typedef pair<int, int> pii; 62 typedef pair<string, int> psi; 63 typedef pair<LL, LL> pll; 64 typedef map<string, int> msi; 65 typedef vector<int> vi; 66 typedef vector<LL> vl; 67 typedef vector<vl> vvl; 68 typedef vector<bool> vb; 69 70 const int maxn = 1100; 71 int dp[maxn][maxn]; 72 char s[maxn]; 73 int n; 74 75 int main() { 76 // FRead(); 77 int T; 78 Rint(T); 79 W(T) { 80 Rs(s); n = strlen(s); 81 Cls(dp); 82 For(k, 2, n+1) { 83 Rep(i, n-k+1) { 84 dp[i][i+k-1] = 0; 85 int j = i + k - 1; 86 if((s[i] == ‘(‘ && s[j] == ‘)‘) || (s[i] == ‘[‘ && s[j] == ‘]‘)) { 87 dp[i][j] = dp[i+1][j-1] + 2; 88 } 89 For(m, i, j) { 90 dp[i][j] = max(dp[i][j], dp[i][m] + dp[m+1][j]); 91 } 92 } 93 } 94 printf("%d\n", n - dp[0][n-1]); 95 } 96 RT 0; 97 }
时间: 2024-10-14 02:08:31