POJ 2955 Brackets (区间DP)

题意：给定一个序列，问你最多有多少个合法的括号。

析：区间DP，dp[i][j] 表示在 第 i 到 第 j 区间内最多有多少个合法的括号。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e2 + 5;
const int mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
char s[maxn];
int dp[maxn][maxn];

bool match(char ch, char sh){
    if(ch == '(' && sh == ')')  return true;
    if(ch == '[' && sh == ']')  return true;
    return false;
}

int main(){
    while(scanf("%s", s) == 1 && s[0] != 'e'){
        n = strlen(s);
        memset(dp, 0, sizeof dp);
        for(int i = n-2; i >= 0; --i)
            for(int j = i+1; j < n; ++j){
                if(match(s[i], s[j]))  dp[i][j] = dp[i+1][j-1] + 2;
                for(int k = i; k < j; ++k)
                    dp[i][j] = Max(dp[i][k]+dp[k+1][j], dp[i][j]);
            }

        printf("%d
", dp[0][n-1]);
    }
    return 0;
}