POJ 题目2955 Brackets（区间dp）

Brackets

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3444 Accepted: 1770

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 … a_n, 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, …, i_mwhere 1 ≤ i₁ < i₂ < … < i_m ≤ n, a_i1a_i2 … a_im 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

66406

Source

Stanford Local 2004

问有多少个括号匹配

ac代码

#include<stdio.h>#include<string.h>#define max(a,b) (a>b?a:b)int dp[1010][1010];char s[110];int jud(int i,int j){if(s[i]=='('&&s[j]==')')return 1;if(s[i]=='['&&s[j]==']')return 1;return 0;}int main(){//char s[110];while(scanf("%s",s)!=EOF){int len,k,i,j,r;memset(dp,0,sizeof(dp));if(strcmp("end",s)==0)break;len=strlen(s);for(i=0;i<len-1;i++){if(jud(i,i+1))dp[i][i+1]=2;elsedp[i][i+1]=0;}for(k=2;k<len;k++){for(i=0;i+k<len;i++){r=k+i;if(jud(i,r))dp[i][r]=dp[i+1][r-1]+2;for(j=i;j<r;j++)//j从i开始{dp[i][r]=max(dp[i][r],dp[i][j]+dp[j+1][r]);}}}printf("%d\n",dp[0][len-1]);}}