POJ 2955 Brackets 区间DP

POJ 2955 Brackets 区间DP

题目链接： POJ 2955 Brackets 

题意：求字符串中匹配的括号数*2，字符串长度<100。

分析：区间DP入门题，dp[i][j] 表示区间[i, j]匹配数 *2

写出状态方程:

if(i, j匹配) 

dp[i][j] = dp[i + 1][j - 1] + 2;

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

#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
using namespace std;

using namespace std;

#define FIN             freopen("input.txt","r",stdin)
#define FOUT            freopen("output.txt","w",stdout)
#define fst             first
#define snd             second

const int MAXN = 100 + 5;

int T, N;
char S[MAXN];
int dp[MAXN][MAXN];
inline bool check(int i, int j) {
    return (S[i] == '(' && S[j] == ')') || (S[i] == '[' && S[j] == ']');
}
int main() {
#ifndef ONLINE_JUDGE
    FIN;
    // FOUT;
#endif // ONLINE_JUDGE
    int len = 0;
    while(~scanf("%s", S)) {
        if(S[0] == 'e') break;
        len = strlen(S);
        memset(dp, 0, sizeof(dp));
        for(int ilen = 1; ilen <= len; ilen++) {
            for(int i = 0, j; i <= len - ilen; i++) {
                j = i + ilen - 1;
                if(check(i, j))
                    dp[i][j] = dp[i + 1][j - 1] + 2;
                for(int k = i; k + 1 <= j; k++) {
                    dp[i][j] = max(dp[i][j], dp[i][k] + dp[k + 1][j]);
                }
            }
        }
        printf("%d\n", dp[0][len - 1]);
    }
    return 0;
}