标签：des style blog class code ext
http://poj.org/problem?id=2955
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 2707 | Accepted: 1403 |
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 msuch 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
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#include<stdio.h>#include<string.h>int dp[100][100];int max(int
x,int y){ if(x>y) return
x; else return
y;}bool
match(char
x,char
y){ if(x==‘[‘&&y==‘]‘) return
true; else
if(x==‘(‘&&y==‘)‘) return
true; else return
false;}int
main(){ int
len,i,j,k,g; char
str[100]; while(gets(str)) { if(str[0]==‘e‘) break; memset(dp,0,sizeof(dp)); len=strlen(str); for(i=0;i<len;i++) { dp[i][i]=0; if(match(str[i],str[i+1])) dp[i][i+1]=2;//初始值。 } for(k=0;k<len;k++) for(i=0;i<len-k;i++) { j=i+k; if(match(str[i],str[j])) dp[i][j]=dp[i+1][j-1]+2; for(g=0;g<k;g++) dp[i][j]=max(dp[i][j],dp[i][i+g]+dp[i+g+1][j]); } printf("%d\n",dp[0][len-1]); } return
0;} |
别人的代码，可供参考。
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#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using
namespace std;const
int maxn=1002;char
s[maxn];int dp[maxn][maxn];int main(){ //freopen("//media/学习/ACM/input.txt","r",stdin); while(scanf("%s",s),s[0]!=‘e‘) { int
i,j,k,n=strlen(s); for(i=0;i<n;i++) for(j=0;j<n;j++) dp[i][j]=0; for(i=n-1;i>=0;i--) { for(j=i+1;j<=n-1;j++) { dp[i][j]=max(dp[i+1][j],dp[i][j-1]); for(k=i+1;k<=j;k++) { if((s[i]==‘(‘&&s[k]==‘)‘)||(s[i]==‘[‘&&s[k]==‘]‘)) dp[i][k]=max(dp[i][k],dp[i+1][k-1]+2); dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]); } // cout<<i<<" "<<j<<" "<<dp[i][j]<<endl; } } cout<<dp[0][n-1]<<endl; } return
0;} |
poj -2955 Brackets,布布扣,bubuko.com
标签：des style blog class code ext
原文地址：http://www.cnblogs.com/cancangood/p/3721013.html
