标签：des style blog io color ar os sp for
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 3340 | Accepted: 1716 |
Description
We give the following inductive definition of a “regular brackets” sequence:
For instance, all of the following character sequences are regular brackets sequences:
(), [], (()), ()[], ()[()]
while the following character sequences are not:
(, ], )(, ([)], ([(]
Given a brackets sequence of characters a1a2 … an, your goal is to find the length of the longest regular brackets sequence that is a subsequence of s. That is, you wish to find the largest m such that for indices i1, i2, …, imwhere 1 ≤ i1 < i2 < … < im ≤ n, ai1ai2 … aim is a regular brackets sequence.
Given the initial sequence ([([]])], the longest regular brackets subsequence is [([])].
Input
The input test file will contain multiple test cases. Each input test case consists of a single line containing only the characters (, ), [, and ]; each input test will have length between 1 and 100, inclusive.
The end-of-file is marked by a line containing the word “end” and should not be processed.
Output
For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.
Sample Input
((())) ()()() ([]]) )[)( ([][][) end
Sample Output
6 6 4 0 6
经典区间DP。
题意：求括号匹配数。
dp[l][r]为区间[l,r]内括号匹配数，dp[l][r]=max(dp[l][r],dp[l+1][k-1]+dp[k+1][r]+2)(s[l]和s[k]匹配)
枚举区间，形式大都一样，3层for循环，看到很多人说区间DP写成记忆化搜索比较容易懂。。orz
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#define ll __int64
#define pp pair<int,int>
using namespace std;
const int INF = 0x3f3f3f3f;
char s[110];
int dp[110][110];
void solve()
{
memset(dp, 0, sizeof(dp));
int len = strlen(s) - 1;
for (int p = 1; p <= len; p++) {
for (int l = 1; l <= len; l++) {
int r = l + p - 1;
if (r > len) {
break;
}
dp[l][r] = dp[l + 1][r];
for (int k = l + 1; k <= r; k++)
if ((s[l] == '(' && s[k] == ')') || (s[l] == '[' && s[k] == ']')) {
dp[l][r] = max(dp[l][r], dp[l + 1][k - 1] + dp[k + 1][r] + 2);
}
}
}
printf("%d\n", dp[1][len]);
}
int main()
{
while (~scanf("%s", s + 1) && s[1] != 'e') {
s[0] = 2;
solve();
}
return 0;
}标签：des style blog io color ar os sp for
原文地址：http://blog.csdn.net/qq_16255321/article/details/40930175
