poj 2955 Brackets 区间dp

题目链接

给一堆括号， 问最多有多少个是匹配的。

依然是区间dp， 直接记忆化搜索就可以。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, a, n) for(int i = a; i<n; i++)
#define ull unsigned long long
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
char s[105];
int dp[105][105];
int judge(int l, int r) {
    if(s[l] == '('&&s[r]==')')
        return 2;
    if(s[l]=='['&&s[r]==']')
        return 2;
    return 0;
}
int dfs(int l, int r) {
    if(~dp[l][r])
        return dp[l][r];
    dp[l][r] = 0;
    if(l>r)
        return 0;
    if(l == r)
        return 0;
    if(r-l == 1) {
        return dp[l][r] = judge(l, r);
    }
    if(judge(l, r))
        dp[l][r] = dfs(l+1, r-1)+2;
    for(int i = l; i<=r; i++) {
        dp[l][r] = max(dp[l][r], dfs(l, i)+dfs(i+1, r));
    }
    return dp[l][r];
}
int main()
{
    while(scanf("%s", s+1)) {
        mem1(dp);
        int len = strlen(s+1);
        if(!strcmp(s+1, "end"))
            break;
        else
            printf("%d
", dfs(1, strlen(s+1)));
    }
    return 0;
}