POJ 2955 Brackets 区间dp

Posted 2020-09-01 stupid_one

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

区间dp的写法还是dfs的好写。

设dp[i][j]表示[i, j]这个区间的合法情况的最大值。

然后想要求[i, j]合法情况的最大值，有两种。

1、如果i和j的搭配的括号，那么dp[i][j] = dp[i + 1][j - 1] + 2;

2、同时还可能是分开区间，合并后才更大，例如：()()

dp[i][j] = max(dp[i][j], dp[i][k] + dp[k  + 1][j]);

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define ios ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
const int maxn = 1e2 + 20;
char str[maxn];
char op[maxn];
int dp[maxn][maxn];
bool match(char ch1, char ch2) {
    if (ch1 == \'(\' && ch2 == \')\') return true;
    if (ch1 == \'[\' && ch2 == \']\') return true;
    return false;
}
int vis[maxn][maxn], DFN;
int dfs(int be, int en) {
    if (be >= en) return 0;
    if (vis[be][en] == DFN) return dp[be][en];
    vis[be][en] = DFN;
    if (match(str[be], str[en])) dp[be][en] = dfs(be + 1, en - 1) + 2;
    for (int k = be; k <= en; ++k) {
        dp[be][en] = max(dp[be][en], dfs(be, k) + dfs(k + 1, en));
    }
    return dp[be][en];
}

void work() {
    DFN++;
    memset(dp, 0, sizeof dp);
    int lenstr = strlen(str + 1);
    cout << dfs(1, lenstr) << endl;
}


int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    op[\'(\'] = \')\';
    op[\')\'] = \'(\';
    op[\'[\'] = \']\';
    op[\']\'] = \'[\';
    while (scanf("%s", str + 1) && str[1] != \'e\') work();
    return 0;
}

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