标签：end nsis reg multi 问题 scanf output other sizeof
题目链接：http://poj.org/problem?id=2955
Time Limit: 1000MS Memory Limit: 65536K
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
题意：
给出一个包含“(”、“)”、“[”、“]”的括号串，求有多少个相匹配的括号；
题解：
区间dp，设dp[i][j]为str[i~j]范围内的答案，即dp[0][len-1]即问题所求答案；
假设我们在求dp[i][j]时，已经知道所有dp[ii][jj]（i<ii<jj<j），那么，做一下两项操作即可：
①先OB一下str[i]和str[j]是不是匹配的，如果是的话，尝试更新dp[i][j]：dp[i][j] = max( dp[i][j] , dp[i+1][j-1]+2 )
②为了保证dp[i][j]的正确性，枚举k=i~j，尝试更新dp[i][j]：dp[i][j] = max( dp[i][j] , dp[i][k]+dp[k][j] )
完美~
AC代码：
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; char str[105]; int dp[105][105]; bool check(char a,char b) { if( (a==‘(‘&&b==‘)‘) || (a==‘[‘&&b==‘]‘) ) return 1; else return 0; } int main() { while(scanf("%s",str)) { if(str[0]==‘e‘) break; int len=strlen(str); memset(dp,0,sizeof(dp)); for(int l=2;l<=len;l++) { for(int i=0,j=i+l-1;j<len;i++,j=i+l-1) { if(check(str[i],str[j])) dp[i][j]=max(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][j]); } } printf("%d\n",dp[0][len-1]); } }
标签：end nsis reg multi 问题 scanf output other sizeof
原文地址：http://www.cnblogs.com/dilthey/p/7875215.html
