POJ 2955 - Brackets(区间DP)

Brackets Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9255   Accepted: 4947 Description We give the following inductive definition of a “regular brackets” sequence: the empty sequence is a regular brackets sequence, if s is a regular brackets sequence, then (s) and [s] are regular brackets sequences, and if a and b are regular brackets sequences, then ab is a regular brackets sequence. no other sequence is 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, …, im where 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 Source Stanford Local 2004

题意：

输出给你的括号匹配最长长度，可以不连续。

POINT：

看代码里的状态转移方程

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <vector>
#include <queue>
using namespace std;
const int maxn = 111;
int dp[maxn][maxn];
int main()
{
char s[maxn];
while(~scanf("%s",s+1)){
if(s[1]=='e') break;
int n=(int)strlen(s)-1;
memset(dp,0,sizeof dp);
for(int len=2;len<=n;len++){
for(int i=1;i<=n-len+1;i++){
int j=i+len-1;
if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']')) dp[i][j]=dp[i+1][j-1]+2;
for(int k=1;k<j;k++)
dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);

}
}
printf("%d\n",dp[1]
);
}
}