http://acm.pku.edu.cn/JudgeOnline/problem?id=1739
基于連通性狀態(tài)壓縮的動(dòng)態(tài)規(guī)劃(名字真長(zhǎng)- -！)其實(shí)不算太難，以前都被它嚇住了。
hyf教了一次后，印象極其深刻，回路、路徑條數(shù)啥的從此再也不用向美國(guó)（霧）進(jìn)口了。
簡(jiǎn)要的說(shuō)：f[i][j][k]表示在（i，j）這個(gè)格子的時(shí)候，m+1個(gè)插頭的情況是k，然后根據(jù)(i,j)左邊和上邊插頭（括號(hào)？）的情況進(jìn)行連接，然后轉(zhuǎn)移到下一個(gè)格子。一個(gè)合法的狀態(tài)是一個(gè)括號(hào)表達(dá)式，有左括號(hào)，右括號(hào)和空格，且左右括號(hào)匹配。一個(gè)左括號(hào)和一個(gè)相應(yīng)的右括號(hào)表示這兩個(gè)地方的路徑是連在一起的。轉(zhuǎn)移的時(shí)候分情況討論。
處理的時(shí)候先把所有合法的狀態(tài)都dfs出來(lái)，轉(zhuǎn)移的時(shí)候就方便了。。
PS：居然驚喜地進(jìn)入了status第一頁(yè)？！。。。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#define MAXN 8
#define BLANK 0
#define LEFT 1
#define RIGHT 2
#define MAXSTATE 174420
#define MAXSTATEAMOUNT 835
#define MAX(a,b) ((a) > (b) ? (a) : (b))
#define ll long long

using namespace std;

int n,m;
char map[MAXN+1][MAXN+1];
int nState = 0;
int State[MAXSTATEAMOUNT+1];
int id[MAXSTATE+1];
int Tx, Ty, FinalState;
ll f[MAXN+1][MAXN+1][MAXSTATEAMOUNT+1];
ll ans;
void Reset(){
     nState = 0, Tx = -1;
     memset(f, 0, sizeof(f));
     ans = 0;
}

bool Init(){
     scanf("%d%d",&n,&m);
     if (!n) return false;
     Reset();
     for (int i = n-1; i>=0; i--){
         scanf("%s", map[i]);
         if (Tx == -1)
         for (int j = m-1; j>=0; j--)
             if (map[i][j] == '.'){
                Tx = i, Ty = j;
                break;
             }
         for (int j = 0; j<m; j++)
             if (map[i][j] == '.') map[i][j] = 0;
             else map[i][j] = 1;
     }
     return true;
}

void dfs(int pos, int r_bracket, int state){
     if (pos < 0){
        State[nState] = state;
        id[state] = nState;
        /*
        printf("%d: ", nState);
        for (int i = 0; i<=m; i++)
            printf("%d ", buffer[i]);
        printf("\n");
        */
        nState ++;
        return;
     }
     if (pos >= r_bracket) // blank
        dfs(pos - 1, r_bracket, (state << 2));
     if (pos > r_bracket) // right bracket
        dfs(pos - 1, r_bracket + 1, (state << 2) | RIGHT);
     if (r_bracket) // left bracket
        dfs(pos - 1, r_bracket - 1, (state << 2) | LEFT);
}

#define MASK 3
#define Get(state, p) (((state) >> (p<<1)) & MASK)

inline bool OK(int i, int j, int state){
     if (Get(state, j) == 1 && Get(state, j+1) == 2 && !(i == Tx && j == Ty)) return false;
     for (int k = 0; k<j; k++)
         if ((map[i][k] || map[i+1][k]) && Get(state, k)) return false;
     if (((j && map[i][j-1]) || (map[i][j])) && Get(state, j)) return false;
     for (int k = j+1; k<=m; k++)
         if (((i && map[i-1][k-1]) || (map[i][k-1])) && Get(state, k)) return false;
     return true;
}

inline int Modify(int state, int p, int v){
    return state - (((state >> (p << 1)) & MASK) << (p << 1)) + (v << (p << 1));
}

inline int FindRight(int p, int state){
    int cnt = 0, t;
    for (int i = p; i<=m; i++){
        t = Get(state,i);
        if (t == 1) cnt++;
        if (t == 2) cnt--;
        if (cnt == 0) return i;
    }
    return -1;
}

inline int FindLeft(int p, int state){
    int cnt = 0, t;
    for (int i = p; i>=0; i--){
        t = Get(state, i);
        if (t == 2) cnt++;
        if (t == 1) cnt--;
        if (cnt == 0) return i;
    }
    return -1;
}

void Solve(){
     dfs(m, 0, 0);
     f[0][0][id[(1 << 2) + (2 << (2 * m))]] = 1;
     FinalState = (1 << (2 * Ty)) + (2 << (2 * (Ty+1)));
     int p, q, tmp, tmp2, state, v, i, j, k;
     ll *a;
     for (i = 0; i < n; i++){
         for (j = 0; j < m; j++){
             a = f[i][j+1];
             for (k = 0; k < nState; k++)
                 if ((v = f[i][j][k])){
                     state = State[k];
                     p = Get(state, j), q = Get(state, j+1);
                     if (p == 0 && q == 0){
                        if (!map[i][j]){
                            tmp = Modify(Modify(state, j, 1), j+1, 2);
                            if (OK(i, j+1, tmp))
                               a[id[tmp]] += v;
                        }else
                             a[k] += v;
                     }else if(p == 0){ // conditions below ensured map[i][j] is empty, because there exists at least one bracket on one side of the grid (i,j)
                           if (OK(i, j+1, state))
                              a[k] += v;
                           tmp = Modify(Modify(state, j, q), j+1, 0);
                           if (OK(i, j+1, tmp))
                              a[id[tmp]] += v;
                     }else if (q == 0){
                           if (OK(i, j+1, state))
                              a[k] += v;
                           tmp = Modify(Modify(state, j, 0), j+1, p);
                           if (OK(i, j+1, tmp))
                              a[id[tmp]] += v;
                     }else{
                         tmp = Modify(Modify(state, j, 0), j+1, 0);
                         if (p == 1 && q == 1){
                               tmp2 = Modify(tmp, FindRight(j+1, state), 1);
                               if (OK(i, j+1, tmp2))
                                  a[id[tmp2]] += v;
                         }else if (p == 2 && q == 2){
                               tmp2 = Modify(tmp, FindLeft(j, state), 2);
                               if (OK(i, j+1, tmp2))
                                  a[id[tmp2]] += v;
                         }else if (p == 1 && q == 2){
                               if (i == Tx && j == Ty && state == FinalState){
                                  printf("%I64d\n", v);
                                  return;
                               }
                         }else if (p == 2 && q == 1){
                               if (OK(i, j+1, tmp))
                                  a[id[tmp]] += v;
                         }
                     }
                 }
         }
         for (int k = 0; k < nState; k++)
             if (Get(State[k], m) == 0 && OK(i+1, 0, tmp = (State[k] << 2)))
                f[i+1][0][id[tmp]] += f[i][m][k];
     }
     printf("%I64d\n", ans);
}

int main(){
    while (Init())
          Solve();
    return 0;
}