LS 24 Bracket sequence(DP)
Bracket sequence
Given string s made up with (
, )
, ?
, count the way to substitude ?
with (
or )
to form regular bracket sequence.
Input
A string s.
(1≤|s|≤1000)
Ouptut
An integer denotes the number of ways modulo (109+7).
Sample input
????
Sample output
2
思路:dp[x][y]字符串长为x时,左括号还有y个未匹配。
#include<iostream>
#include<cstring>
using namespace std;
const int mm=1009;
const int mod=1e9+7;
char s[mm];
int dp[2][mm];
int main()
{while(cin>>s){int len=strlen(s);memset(dp,0,sizeof(dp));dp[0][0]=1;int z=0;for(int i=1;i<=len;i++){if(s[i-1]=='('){for(int j=1;j<=i;j++)dp[z^1][j]=dp[z][j-1];dp[z^1][0]=0;}else if(s[i-1]==')'){for(int j=0;j<i;j++)dp[z^1][j]=dp[z][j+1];}else{for(int j=1;j<=i;j++)dp[z^1][j]=dp[z][j-1];dp[z^1][0]=0;for(int j=0;j<i;j++)dp[z^1][j]=(dp[z][j+1]+dp[z^1][j])%mod;}z^=1;}cout<<dp[z][0]<<"\n";}
}
转载于:https://www.cnblogs.com/nealgavin/archive/2013/03/13/3206130.html
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