POJ 2955 Brackets（区间DP）

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http://poj.org/problem?id=2955.

小括号和中括号分别两两匹配。

用记忆化搜索扫描区间。左右边界为 l, r。每次dfs，我们扫描l到r的所有中间点，即从l+1到r-1，从而更新dp[l][r]的最大值。

转移方程：dp[l][r] = max(dp[l][r], 2 + dfs(l + 1, m - 1) + dfs(m + 1, r))。

#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdio>#include <vector>#include <string>#include <queue>#include <stack>#include <cmath>#include <set>#include <map>using namespace std;typedef long long LL;#define mem(a, n) memset(a, n, sizeof(a))#define ALL(v) v.begin(), v.end()#define si(a) scanf("%d", &a)#define sii(a, b) scanf("%d%d", &a, &b)#define siii(a, b, c) scanf("%d%d%d", &a, &b, &c)#define pb push_back#define eps 1e-8const int inf = 0x3f3f3f3f, N = 1e2 + 5, MOD = 1e9 + 7;int T, cas = 0;int n, m;char s[N];int dp[N][N];bool check(int i, int j) {return (s[i] == '(' && s[j] == ')') || (s[i] == '[' && s[j] == ']');}int dfs(int l, int r) {int& ret = dp[l][r];if(ret != -1) return ret;if(l >= r) return ret = 0;if(r == l + 1) return ret = check(l, r) ? 2 : 0;ret = dfs(l + 1, r);for(int m = l + 1; m <= r; m ++)if(check(l, m)) ret = max(ret, 2 + dfs(l + 1, m - 1) + dfs(m + 1, r));return ret;}int main(){#ifdef LOCAL    freopen("/Users/apple/input.txt", "r", stdin);//freopen("/Users/apple/out.txt", "w", stdout);#endif    while(scanf("%s", s) != EOF) {    if(s[0] == 'e') break;    mem(dp, -1);    int len = strlen(s);    printf("%d\n", dfs(0, len - 1));    }        return 0;}

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