gym100825G. Tray Bien(轮廓线DP)

题意:3 * N的格子 有一些点是坏的 用1X1和1X2的砖铺有多少种方法

题解:重新学了下轮廓线 写的很舒服

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int n, m;
int vis[][];
ll dp[][ << ];

void dfs(int num, int i, int state, int nex)
{
    if(num == )
    {
        dp[i + ][nex] += dp[i][state];
        return;
    }

    if(vis[i][num + ] || (state & ( << num))) dfs(num + , i, state, nex);
    else
    {
        dfs(num + , i, state, nex); //填1x1
        if(!vis[i + ][num + ] && !(nex & ( << num)))
            dfs(num + , i, state, nex | ( << num));   //竖着填1X2
        if(num +  <=  && !vis[i][num + ] && !(state & ( << (num + ))))
            dfs(num + , i, state, nex); // 横着填1X2
    }
}

int main()
{
    scanf("%d%d", &n, &m);
    for(int i = ; i <= m; i++)
    {
        double x, y;
        scanf("%lf%lf", &x, &y);
        vis[(int)x + ][(int)y + ] = ;
    }

    dp[][] = 1LL;
    for(int i = ; i <= n; i++)
        for(int st = ; st < ; st++)
            if(dp[i][st] > )
            {
                bool f = true;
                for(int k = ; k < ; k++)
                {
                    if(st & ( << k) && vis[i][k + ])
                    {
                        f = false;
                        break;
                    }
                }
                if(!f) continue;
                dfs(, i, st, );
            }

    printf("%lld\n", dp[n + ][]);
    return ;
}