Poj 2955 (interval DP)

Question Link: Http://poj.org/problem? Id = 2955

Theme: Matching brackets. The number of symmetric parentheses matches + 2. Maximum number of matches.

Solutions:

It looks like a range problem.

DP border: none. When the interval is 0, the default value is memset 0.

For DP [I] [J], if I and j match, it is not difficult to have DP [I] [J] = DP [I + 1] [J-1] + 2.

Then enumeration does not belong to the midpoint at both ends, DP [I] [J] = max (DP [I] [J], DP [I] [k] + dp [k] [J]), and merge the results of the two intervals.

 

#include "cstdio"#include "string"#include "cstring"#include "iostream"using namespace std;bool check(char a,int b){    if(a==‘(‘&&b==‘)‘) return true;    else if(a==‘[‘&&b==‘]‘) return true;    else return false;}int dp[105][105];int main(){    //freopen("in.txt","r",stdin);    string str;    while(cin>>str&&str!="end")    {        int n=str.size();        for(int p=1;p<n;p++)        {            for(int i=0;i<n-p;i++)            {                int j=i+p;                if(check(str[i],str[j])) dp[i][j]=dp[i+1][j-1]+2;                for(int k=i+1;k<j;k++)                    dp[i][j]=max(dp[i][j],dp[i][k]+dp[k][j]);            }        }        printf("%d\n",dp[0][n-1]);        memset(dp,0,sizeof(dp));    }}

 

Poj 2955 (interval DP)