題意：給定一個01矩陣，問你能畫出幾條回路，使得包含所有的1。

析：一個插頭DP，dp[i][j][s] 表示轉移到 (i, j) 這個格子，狀態為 s 時的方案數，然後逐格遞推。對於每個格子要麽有0個插頭要麽有2個。

代碼如下：

#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 = 11 + 10;
const ULL mod = 10007;
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;
}

int a[maxn][maxn];

LL dp[2][1<<12];

int main(){
  int T;  cin >> T;
  for(int kase = 1; kase <= T; ++kase){
    scanf("%d %d", &n, &m);
    for(int i = 1; i <= n; ++i)
      for(int j = 1; j <= m; ++j)
        scanf("%d", &a[i][j]);
    int cur = 0;
    ms(dp[cur], 0);
    dp[cur][0] = 1;
    int all = 1<<m+1;
    for(int i = 1; i <= n; ++i){
      cur ^= 1; ms(dp[cur], 0);
      for(int j = 0; j < (1<<m); ++j)
        dp[cur][j<<1] = dp[cur^1][j];
      for(int j = 1; j <= m; ++j){
        cur ^= 1;
        for(int k = 0; k < all; ++k){
          int up = 1<<j;
          int le = 1<<j-1;
          if(a[i][j]){
            dp[cur][k] = dp[cur^1][k^up^le];
            if(k&up && k&le)  continue;
            if(!(k&up) && !(k&le))  continue;
            dp[cur][k] += dp[cur^1][k];
          }
          else{
            if(!(k&up) && !(k&le))  dp[cur][k] = dp[cur^1][k];
            else  dp[cur][k] = 0;
          }
        }
      }
    }
    printf("Case %d: There are %I64d ways to eat the trees.\n", kase, dp[cur][0]);
  }
  return 0;
}

HDU 1693 Eat the Trees (插頭DP)