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[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\n", dp[n + 1][0]); return 0; }