FZU 1977 Pandora adventure (DP)

题意：给定一个图，X表示不能走，O表示必须要走，*表示可走可不走，问你多少种走的法，使得形成一个回路。

析：

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
//#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 10 + 10;
const int maxm = 1e5 + 10;
const int mod = 50007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

struct Hash{
  int head[mod], next[maxm], sz;
  LL state[maxm], f[maxm];
  void clear(){  sz = 0;  ms(head, -1);  }
  void push(LL st, LL ans){
    int h = st % mod;
    for(int i = head[h]; ~i; i = next[i])
      if(st == state[i]){
        f[i] += ans;
        return ;
      }
    f[sz] = ans;
    state[sz] = st;
    next[sz] = head[h];
    head[h] = sz++;
  }
};
Hash dp[2];
int isend;  // is or not form loop
int code[maxn], ch[maxn];
int a[maxn][maxn];  // 1 must go 2 go or not
char s[maxn];

void decode(int m, LL st){
  for(int i = m; i >= 0; --i){
    code[i] = st&7;
    st >>= 3;
  }
  isend = st&1;  // the highest bit
}

LL encode(int m){
  LL st = isend;
  int cnt = 1;  ms(ch, -1);
  ch[0] = 0;
  for(int i = 0; i <= m; ++i){
    if(ch[code[i]] == -1)  ch[code[i]] = cnt++;
    st <<= 3;
    st |= ch[code[i]];
  }
  return st;
}

void shift(int m){
  for(int i = m; i; --i)  code[i] = code[i-1];
  code[0] = 0;
}

void dpblank(int i, int j, int cur){
  for(int k = 0; k < dp[cur].sz; ++k){
    decode(m, dp[cur].state[k]);
    int left = code[j-1];
    int up = code[j];
    if(isend){  // already form loop
      if(up || left || a[i][j] == 1)  continue;
      code[j] = code[j-1] = 0;
      if(j == m)  shift(m);
      dp[cur^1].push(encode(m), dp[cur].f[k]);
      continue;
    }
    if(up && left){
      if(up == left){  // last grid
        code[j-1] = code[j] = 0;
        isend = 1;
        if(j == m)  shift(m);
        dp[cur^1].push(encode(m), dp[cur].f[k]);
      }
      else{
        code[j] = code[j-1] = 0;
        for(int i = 0; i <= m; ++i)
          if(code[i] == up)  code[i] = left;
        if(j == m)  shift(m);
        dp[cur^1].push(encode(m), dp[cur].f[k]);
      }
    }
    else if(up || left){
      int t = max(up, left);
      if(a[i][j+1]){
        code[j] = t;
        code[j-1] = 0;
        dp[cur^1].push(encode(m), dp[cur].f[k]);
      }
      if(a[i+1][j]){
        code[j] = 0;
        code[j-1] = t;
        if(j == m)  shift(m);
        dp[cur^1].push(encode(m), dp[cur].f[k]);
      }
    }
    else{
      if(a[i+1][j] && a[i][j+1]){
        code[j] = code[j-1] = 14;
        dp[cur^1].push(encode(m), dp[cur].f[k]);
      }
      if(a[i][j] == 2){
        code[j] = code[j-1] = 0;
        if(j == m)  shift(m);
        dp[cur^1].push(encode(m), dp[cur].f[k]);
      }
    }
  }
}

void dpblock(int i, int j, int cur){
  for(int k = 0; k < dp[cur].sz; ++k){
    decode(m, dp[cur].state[k]);
    code[j] = code[j-1] = 0;
    if(j == m)  shift(m);
    dp[cur^1].push(encode(m), dp[cur].f[k]);
  }
}

int main(){
  int T;  cin >> T;
  for(int kase = 1; kase <= T; ++kase){
    scanf("%d %d", &n, &m);
    ms(a, 0);
    for(int i = 1; i <= n; ++i){
      scanf("%s", s + 1);
      for(int j = 1; j <= m; ++j){
        if(s[j] == 'O')  a[i][j] = 1;  // must go
        else if(s[j] == '*')  a[i][j] = 2;  // not need go
      }
    }
    int cur = 0;
    dp[cur].cl; dp[cur].push(0, 1);
    for(int i = 1; i <= n; ++i)
      for(int j = 1; j <= m; ++j){
        dp[cur^1].cl;
        if(a[i][j])  dpblank(i, j, cur);
        else dpblock(i, j, cur);
        cur ^= 1;
      }
    LL ans = 0;
    for(int i = 0; i < dp[cur].sz; ++i)
      ans += dp[cur].f[i];
    printf("Case %d: %I64d\n", kase, ans);
  }
  return 0;
}