CF 342D

CF 342D

题意：给出一个N*3的方块，要求用2*1的方砖填到只剩下一个1*1的空位，要求空位四周至少有一个方砖用长度为1的面对着它，求方案数。好恶心，英语渣打完一遍才发现看错题了。

状态压缩+容斥原理。

如果只要求填满，那么就很简单了，随便找个模板套上去就差不多了，但是后面一个要求有点麻烦，首先假设空位的左边有一块符合条件的方砖，那么就把他自身跟左边两个格子都填上不能放置的标志，然后其他地方就用简单状态压缩去做，右面，上下两面都是这样处理，但是这样求出来的方案数有重复，所以用容斥原理求出同时有两个/三个/四个方向都有符合条件的方砖，再做几次状态压缩，搞定。

#include <cstdio>#include <cstring>#include <cmath>#include <queue>#include <stack>#include <vector>#include <string>#include <map>#include <set>#include <sstream>#include <iostream>#include <algorithm>#include<cstdlib>#include<queue>using namespace std;#define N 10005#define L(x) x<<1#define R(x) x<<1|1#define M(x,y) (x + y)>>1#define MOD 1000000007#define MODD 1000000006#define inf 0x7fffffff#define llinf 0x7fffffffffffffff#define LL long longLL s[N][5];LL dp[N][10];LL bin[10][4] = {{0,0,0,0},{0,1,0,0},{0,0,1,0},{0,1,1,0},{0,0,0,1},{0,1,0,1},{0,0,1,1},{0,1,1,1}};LL b[4] = {0,1,2,4};LL i_e[10],num;LL comb[5];void dfs(LL x,LL y,LL th,LL len){if(th == 4){dp[len][y] = (dp[len][y] + dp[len - 1][x])%MOD;return ;}if(bin[y][th] == 1 && bin[x][th] == 0)return ;if(bin[y][th] == 1 && bin[x][th] == 1)dfs(x,y,th + 1,len);if(bin[y][th] == 0 && bin[x][th] == 0)dfs(x,y + b[th],th + 1,len);if(bin[y][th] == 0 && bin[x][th] == 1)dfs(x,y,th + 1,len);if(th > 1 && bin[y][th] + bin[y][th - 1] == 0 && bin[x][th] + bin[x][th - 1] == 2)dfs(x,y + b[th] + b[th - 1],th + 1,len);}LL cntdp(LL n){memset(dp,0,sizeof(dp));dp[0][7] = 1;for(LL i = 1;i <= n;i++){for(LL j = 0;j < 8;j++){if(dp[i - 1][j] == 0)continue;LL cnt = 0;for(LL k = 1;k <= 3;k++){if(s[i][k] != 0){cnt += b[k];}}dfs(j,cnt,1,i);}}return dp[n][7];}void updata(LL n,LL x,LL y){if(n == 1){s[y - 1][x] = s[y - 2][x] = 1;}else if(n == 2){s[y + 2][x] = s[y + 1][x] = 1;}else if(n == 3){s[y][1] = s[y][2] = 1;}else{s[y][2] = s[y][3] = 1;}}void dowmdata(LL n,LL x,LL y){if(n == 1){s[y - 1][x] = s[y - 2][x] = 0;}else if(n == 2){s[y + 2][x] = s[y + 1][x] = 0;}else if(n == 3){s[y][1] = s[y][2] = 0;}else{s[y][2] = s[y][3] = 0;}}LL dfs1(LL n,LL m,LL x,LL y,LL number){if(m > n){for(LL i = 1;i <= n;i++){updata(comb[i],x,y);}LL sum = cntdp(number);for(LL i = 1;i <= n;i++){dowmdata(comb[i],x,y);}return sum;}LL sum = 0;for(LL i = 1;i <= num;i++){if(comb[m - 1] >= i_e[i] && m != 1)continue;comb[m] = i_e[i];sum = (sum + dfs1(n,m + 1,x,y,number))%MOD;}return sum;}int main(){LL i,j,k,l;LL n,x,y;while(cin>>n){for(i = 1;i <= 3;i++){for(j = 1;j <= n;j++){char c;cin>>c;if(c == '.')s[j][i] = 0;else if(c == 'X')s[j][i] = 1;else{y = j;x = i;s[j][i] = 1;}}}num = 0;LL ans = 0;if(y > 2 && s[y - 2][x] + s[y - 1][x] == 0){i_e[++num] = 1;}if(y < n - 2 && s[y + 2][x] + s[y + 1][x] == 0){i_e[++num] = 2;}if(x == 3 && s[y][1] + s[y][2] == 0){i_e[++num] = 3;}if(x == 1 && s[y][2] + s[y][3] == 0){i_e[++num] = 4;}for(i = 1;i <= num;i++){ans = (ans + dfs1(i,1,x,y,n)*(i%2 == 0?-1:1))%MOD;}ans = (ans + MOD)%MOD;cout<<ans<<endl;}return 0;}