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最新推荐文章于 2022-07-11 12:07:29 发布

原创最新推荐文章于 2022-07-11 12:07:29 发布 · 195 阅读

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本文探讨了一个复杂的C++程序，该程序利用动态规划解决括号匹配问题，通过对字符串中的括号进行深度优先搜索，计算各种匹配情况下的状态转移方程，最终求得所有可能的匹配组合数。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mem(a, b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define DBG printf("this is a input\n")
string s;
int match[800];
ll dp[700][700][5][5];
stack<int>st;
const int mod = 1000000007;
void dfs(int l , int r)
{
    //cout<<l<<" "<<r<<endl;
    if(r == l+1)
        dp[l][r][0][1] = dp[l][r][0][2] = dp[l][r][1][0] = dp[l][r][2][0] = 1;
    else if(match[l] == r)
    {
        dfs(l+1,r-1);
        for(int i = 0 ; i < 3 ; i ++)
        {
            for (int j = 0; j < 3; j++)
            {
                if(i != 1) dp[l][r][1][0] = (dp[l][r][1][0] + dp[l+1][r-1][i][j])%mod;
                if(i != 2) dp[l][r][2][0] = (dp[l][r][2][0] + dp[l+1][r-1][i][j])%mod;
                if(j != 1) dp[l][r][0][1] = (dp[l][r][0][1] + dp[l+1][r-1][i][j])%mod;
                if(j != 2) dp[l][r][0][2] = (dp[l][r][0][2] + dp[l+1][r-1][i][j])%mod;
            }
        }
    }
    else
    {

        dfs(l,match[l]);
        dfs(match[l]+1,r);
        for(int i = 0 ; i < 3 ; i ++)
        {
            for(int j = 0 ; j < 3 ; j ++)
            {
                for(int x = 0 ; x < 3 ; x ++)
                {
                    for(int y = 0 ; y < 3 ; y ++)
                    {
                        if(!((j==1 && x == 1) || (j == 2 && x == 2)))
                            dp[l][r][i][y] = (dp[l][r][i][y] + (dp[l][match[l]][i][j] * dp[match[l] + 1][r][x][y])%mod)%mod;
                        /*
                        //cout<<i<<" "<<j<<" "<<x<<" "<<y<<" "<<dp[l][match[l]][i][j]<<" "<<dp[match[l]+1][r][x][y]<<endl;
                        if(((i == 0 && j != 0) || (i != 0 && j == 0)) && ((x == 0 && y != 0) || (x != 0 && y == 0)))
                        {
                            //cout<<i<<" "<<j<<" "<<x<<" "<<y<<" "<<dp[l][match[l]][i][j]<<" "<<dp[match[l]+1][r][x][y]<<endl;
                            if(j == 0 && x == 0)
                            {
                                cout <<l<<" "<<r<<" "<< i << " " << j << " " << x << " " << y << " " << dp[l][match[l]][i][j] << " " << dp[match[l] + 1][r][x][y] << endl;
                                dp[l][r][i][y] = (dp[l][r][i][y] + (dp[l][match[l]][i][j] * dp[match[l] + 1][r][x][y])%mod)%mod;
                            }
                            if(j != x)
                            {
                                cout <<l<<" "<<r<<" "<< i << " " << j << " " << x << " " << y << " " << dp[l][match[l]][i][j] << " " << dp[match[l] + 1][r][x][y] << endl;
                                dp[l][r][i][y] = (dp[l][r][i][y] + (dp[l][match[l]][i][j] * dp[match[l] + 1][r][x][y])%mod)%mod;
                            }
                            //cout << "dp: " << l << " " << r << " " << i << " " << j << " " << dp[l][r][i][y] << endl;
                        }
                         */
                    }
                }
            }
        }
    }
}
int main(void)
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin>>s;
    for(int i = 0 ; i < s.length() ; i ++)
    {
        if(s[i] == '(')
            st.push(i);
        else
        {
            match[st.top()] = i;
            st.pop();
        }
    }
    dfs(0,s.length()-1);
    ll ans = 0;
    /*
    for(int i = 0 ; i <= 2 ; i ++)
        for(int j = 0 ; j <= 2 ; j ++)
            cout<<i<<" "<<j<<" "<<dp[2][s.length()-1][i][j]<<endl;
    cout<<"!!!!!!!"<<endl;
     */
    for(int i = 0 ; i <= 2 ; i ++)
        for(int j = 0 ; j <= 2 ; j ++)
            ans = (ans + dp[0][s.length()-1][i][j])%mod;
    cout<<ans<<endl;
}