gym100825G. Tray Bien(轮廓线DP)

Posted 2021-01-10 lwqq3

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[30][5];
ll dp[25][1 << 3];

void dfs(int num, int i, int state, int nex)
{
    if(num == 3)
    {
        dp[i + 1][nex] += dp[i][state];
        return;
    }

    if(vis[i][num + 1] || (state & (1 << num))) dfs(num + 1, i, state, nex);
    else
    {
        dfs(num + 1, i, state, nex); //填1x1
        if(!vis[i + 1][num + 1] && !(nex & (1 << num)))
            dfs(num + 1, i, state, nex | (1 << num));   //竖着填1X2
        if(num + 2 <= 3 && !vis[i][num + 2] && !(state & (1 << (num + 1))))
            dfs(num + 2, i, state, nex); // 横着填1X2
    }
}

int main()
{
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; i++)
    {
        double x, y;
        scanf("%lf%lf", &x, &y);
        vis[(int)x + 1][(int)y + 1] = 1;
    }

    dp[1][0] = 1LL;
    for(int i = 1; i <= n; i++)
        for(int st = 0; st < 8; st++)
            if(dp[i][st] > 0)
            {
                bool f = true;
                for(int k = 0; k < 3; k++)
                {
                    if(st & (1 << k) && vis[i][k + 1])
                    {
                        f = false;
                        break;
                    }
                }
                if(!f) continue;
                dfs(0, i, st, 0);
            }

    printf("%lld
", dp[n + 1][0]);
    return 0;
}

View Code