SCU2016-02 Q题区间dp

Analyse:
经典的区间dp，首先从1开始枚举区间的长度L，然后枚举左端点i，右端点j为i + L。
然后当前区间的值要么通过小区间扩展而来，要么通过小区间合并而来。
Get:

/**********************jibancanyang************************** *Author*        :jibancanyang *Created Time*  : 三  7/ 6 14:07:06 2016**Problem**:**Code**:***********************[e-mail]**********************/#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <cstdlib>#include <ctime>#include <stack>using namespace std;typedef pair<int, int> pii;typedef long long ll;typedef unsigned long long ull;typedef vector<int> vi;#define pr(x) cout << #x << ": " << x << "  " #define pl(x) cout << #x << ": " << x << endl;#define pri(a) printf("%d\n",(a))#define xx first#define yy second#define sa(n) scanf("%d", &(n))#define sal(n) scanf("%lld", &(n))#define sai(n) scanf("%I64d", &(n))#define vep(c) for(decltype((c).begin() ) it = (c).begin(); it != (c).end(); it++) const int mod = int(1e9) + 7, INF = 0x3f3f3f3f;const int maxn = 1e5 + 13;int dp[111][111];int main(void){#ifdef LOCAL    freopen("in.txt", "r", stdin);    //freopen("out.txt", "w", stdout);#endif    string str;    while (cin >> str, str != "end") {        int n = str.size();        memset(dp, 0, sizeof(dp));        for (int L = 1; L <= n; L++) {            for (int i = 0; i < n; i++) {                int j = i + L;                if (j >= n) break;                if ((str[i] == '[' && str[j] == ']') || (str[i] == '(' && str[j] == ')')) {                    dp[i][j] = max((j - i > 2 ? dp[i + 1][j - 1]  : 0) + 2, dp[i][j]);                }                for (int k = i ; k < j; k++) dp[i][j] = max(dp[i][j], dp[i][k] + dp[k + 1][j]);            }        }        pri(dp[0][n - 1]);    }    return 0;}