bzoj2461: [BeiJing2011]符环

再做水题就没救了

考虑DP。。。我们把正反面一起弄

fi,j,k,u表示第几个位置，正面多了多少左括号，背面多了多少没办法消的右括号，背面结尾的左括号数

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;

LL f[][][][];
char ss[];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",ss+); int n=strlen(ss+);

        memset(f,,sizeof(f));
        if(ss[]=='S')f[][][][]=;
        else f[][][][]=;
        for(int i=;i<n;i++)
            for(int j=;j<=i;j++)
                for(int k=;k<=i;k++)
                    for(int u=;u+k<=i;u++)
                        if(f[i][j][k][u]>)
                        {
                            if(ss[i+]=='S')
                            {
                                f[i+][j+][k][u+]+=f[i][j][k][u];
                                if(j>)
                                {
                                    if(u>)f[i+][j-][k][u-]+=f[i][j][k][u];
                                    else f[i+][j-][k+][u]+=f[i][j][k][u];
                                }
                            }
                            else
                            {
                                if(u>)f[i+][j+][k][u-]+=f[i][j][k][u];
                                else f[i+][j+][k+][u]+=f[i][j][k][u];

                                if(j>)f[i+][j-][k][u+]+=f[i][j][k][u];
                            }
                     }
        LL ans=;
        for(int j=;j<=n;j++)
            ans+=f[n][j][j][];
        printf("%lld\n",ans);
    }

    return ;
}