标签：dp
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, …, 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
dp[i][j]表示在区间i到j匹配的最大数。
#include<stdio.h> #include<string.h> int max(int a,int b) { return a>b?a:b; } int main() { int dp[105][105]; char str[105]; while(scanf("%s",str)>0&&strcmp(str,"end")!=0) { int len=strlen(str); memset(dp,0,sizeof(dp)); for(int l=1;l<len;l++)//所求区间头尾相差长度 for(int i=0;i<len-l;i++)//区间起始位置 { int j=l+i;//区间尾部位置 dp[i][j]=dp[i+1][j];//当第i个在这段区间内没有匹配的时 for(int k=i+1;k<=j;k++)//当第i个与第k个位置匹配上时，状态转移如下 if(str[i]=='('&&str[k]==')'||str[i]=='['&&str[k]==']') dp[i][j]=max(dp[i][j],dp[i+1][k-1]+dp[k+1][j]+2); } printf("%d\n",dp[0][len-1]); } }
标签：dp
原文地址：http://blog.csdn.net/u010372095/article/details/37969537
