（区间dp）poj 2955 Brackets

题目
poj2955

题意：
给出字符串，求字符串中有（可以不连续）有多少对“ () " 和 “ [] ”

思路：
在区间 [ i , j ] 内，
如果a [ i ] = ‘(’ 并且a [ j ] = ‘)’（或a [ i ] = ‘[’ 并且a [ j ] = ‘]’ ），则
$$dp[i][j]=max(\underset{k=i}{\overset{j-1}{\max }}dp[i][k]+dp[k+1][j],dp[i+1][j-1])$$
否则，
$$dp[i][j]=\underset{k=i}{\overset{j-1}{\max }}dp[i][k]+dp[k+1][j]$$

代码

#include <iostream>
#include <algorithm>
#include <cstring>
#define DEBUG freopen("_in.txt", "r", stdin); freopen("_out1.txt", "w", stdout);
using namespace std;
const int MAXN = 110;
char a[MAXN], b[MAXN], t[] = "([])";  //a:输入的字符串，b:字符串每个字符对应的更改字符
int dp[MAXN][MAXN], need[200];
void solve(){
    
    
	int lena = strlen(a);
	for (int i = 0; i < lena; i++)  //记录字符串每个字符对应的字符
		b[i] = t[need[a[i]]];
	for (int i = 0; i < lena; i++)
		dp[i][i] = 0;
	for (int len = 1; len < lena; len++)
		for (int i = 0; i < lena - len; i++){
    
    
			int j = i + len;
			dp[i][j] = 0;
			if (a[i] == b[j] && a[i] < a[j])  //()和[]
				dp[i][j] = dp[i+1][j-1] + 2;
			for (int k = i; k < j; k++)
				dp[i][j] = max(dp[i][j], dp[i][k] + dp[k+1][j]);
		}
	printf("%d\n", dp[0][lena-1]);
}
int main(){
    
    
	for (int i = 0; i < 4; i++)
		need[t[i]] = 3 - i;
	while (~scanf("%s", a) && strcmp(a, "end") != 0)
		solve();
	return 0;
}