poj2955 Brackets

题意：就是数有一个字符串中有多少括号匹配：① （）算两种，② [ ] 算两种

题解: 和 LightOj 那道题一样

　　  F[ i ][ j ] = max(F[ i + 1][ j ], F[ i + 1][ k - 1] + F[ k ][ j ] + 2) {i + 1 <= k <= j};

CODE:

/*
Author: JDD
PROG: poj2955 Brackets
DATE: 2015.10.8
*/

#include <cstdio>
#include <cstring>
#define REP(i, s, n) for(int i = s; i <= n; i ++)
#define REP_(i, s, n) for(int i = n; i >= s; i --)
#define MAX_N 105

using namespace std;

char s[MAX_N];
int n, a[MAX_N], F[MAX_N][MAX_N];

#define max(a, b) (a > b ? a : b)

void doit()
{
    memset(F, 0, sizeof(MAX_N));
    REP_(i, 1, n - 1) REP(j, i + 1, n){
        F[i][j] = F[i + 1][j];
        REP(k, i + 1, j)
            if((s[i] == ‘(‘ && s[k] == ‘)‘) || (s[i] == ‘[‘ && s[k] == ‘]‘)) F[i][j] = max(F[i][j], F[i + 1][k - 1] + F[k][j] + 2);
    }
    printf("%d\n", F[1][n]);
}

void init()
{
    while(scanf("%s", s + 1) != EOF){
        n = strlen(s + 1);
        if(n == 3 && s[1] == ‘e‘ && s[2] == ‘n‘ && s[3] == ‘d‘) break;
        doit();
    }
}

int main()
{
    init();
    return 0;
}