题意:就是上一道formula的基础上限定某些格子只能竖着通过,某些只能横着通过。
还是括号表示。这个转移的时候特判一下就好了,具体实现基本和上一题一样。
#include<cstdio> #include<cstring> #include<cstdlib> #include<cassert> #include<iostream> using namespace std; #define LL long long #define clr(a) memset(a,0,sizeof a) #define rep(i,a,b) for(int i=a;i<=b;++i) #define erp(i,a,b) for(int i=a;i>=b;--i) #define getbit(x,y) (((x) >> ((y)<<1)) & 3) #define bit(x,y) ((x)<<((y)<<1)) #define clrbit(x,i,j) ((x) & (~(3<<((i)<<1))) & (~(3<<((j)<<1)))) const int MAXS = 1000000, mo = 100003; int n, m, ex, ey; char a[15][15]; struct Node { int s, nxt; LL val; }; struct Hash { Node e[MAXS]; int adj[mo], ec; void init() { memset(adj, -1, sizeof adj); ec = 0; } void push(int s, LL v) { int ha = s%mo; for (int i = adj[ha]; ~i; i=e[i].nxt) if (e[i].s == s) { e[i].val += v; return; } e[ec].val = v, e[ec].s = s; e[ec].nxt = adj[ha]; adj[ha] = ec++; } } dp[2]; inline int FindL(int st, int x) { int cnt = 1, s; erp(i, x-1, 0) { s = (st >> (i<<1)) & 3; if (s == 2) cnt++; else if (s == 1) cnt--; if (!cnt) return i; } return -1; } inline int FindR(int st, int x) { int cnt = 1, s; rep(i, x+1, m) { s = (st >> (i<<1)) & 3; if (s == 1) cnt++; else if (s == 2) cnt--; if (!cnt) return i; } return -1; } void work1(int i, int j, int cur) // '.' { dp[cur].init(); rep(k, 0, dp[cur^1].ec-1) { int L = getbit(dp[cur^1].e[k].s, j-1); int U = getbit(dp[cur^1].e[k].s, j); int s = clrbit(dp[cur^1].e[k].s, j-1, j); LL las = dp[cur^1].e[k].val; if (!L && !U) { if (i<n && j<m) dp[cur].push(s|bit(1, j-1)|bit(2, j), las); } else if (!L) { if (i < n) dp[cur].push(s|bit(U, j-1), las); if (j < m) dp[cur].push(s|bit(U, j), las); } else if (!U) { if (i < n) dp[cur].push(s|bit(L, j-1), las); if (j < m) dp[cur].push(s|bit(L, j), las); } else if (L==1 && U==1) dp[cur].push(s^bit(3, FindR(s, j)), las); else if (L==2 && U==2) dp[cur].push(s^bit(3, FindL(s, j-1)), las); else if (L==2 && U==1) dp[cur].push(s, las); else if (i==ex && j==ey) dp[cur].push(s, las); } } void work2(int i, int j, int cur) { dp[cur].init(); rep(k, 0, dp[cur^1].ec-1) { int L = getbit(dp[cur^1].e[k].s, j-1); int U = getbit(dp[cur^1].e[k].s, j); int s = clrbit(dp[cur^1].e[k].s, j-1, j); LL las = dp[cur^1].e[k].val; if (!L && !U) dp[cur].push(s, las); } } void worklr(int i, int j, int cur) { dp[cur].init(); rep(k, 0, dp[cur^1].ec-1) { int L = getbit(dp[cur^1].e[k].s, j-1); int U = getbit(dp[cur^1].e[k].s, j); int s = clrbit(dp[cur^1].e[k].s, j-1, j); LL las = dp[cur^1].e[k].val; if (!U && L) { if (j < m) dp[cur].push(s|bit(L, j), las); } } } void workud(int i, int j, int cur) { dp[cur].init(); rep(k, 0, dp[cur^1].ec-1) { int L = getbit(dp[cur^1].e[k].s, j-1); int U = getbit(dp[cur^1].e[k].s, j); int s = clrbit(dp[cur^1].e[k].s, j-1, j); LL las = dp[cur^1].e[k].val; if (!L && U) { if (i < n) dp[cur].push(s|bit(U, j-1), las); } } } LL solve() { dp[0].init(), dp[0].push(0, 1); int cur = 0; rep(i, 1, n) { rep(k, 0, dp[cur].ec-1) dp[cur].e[k].s <<= 2; rep(j, 1, m) { cur ^= 1; if (a[i][j]=='.') work1(i, j, cur); else if (a[i][j]=='#') work2(i, j, cur); else if (a[i][j]=='-') worklr(i, j, cur); else workud(i, j, cur); } } rep(k, 0, dp[cur].ec-1) if (!dp[cur].e[k].s) return dp[cur].e[k].val; return 0; } int main() { scanf("%d%d", &n, &m); rep(i, 1, n) { scanf("%s", a[i]+1); rep(j, 1, m) { if (a[i][j]=='.') ex = i, ey = j; if (a[i][j]=='-'&&(j==1||j==m)) { puts("0"); return 0; } if (a[i][j]=='|'&&(i==1||i==n)) { puts("0"); return 0; } } } cout << solve() << '\n'; return 0; }