HDU 1693 Eat the Trees 插头DP

求一个棋盘不定量条回路覆盖方案数。

是不是插头dp的入门题呢。。

对于正在决策的格子有几种情况。

1. 障碍

    1) 有插头：显然方案数为0。

    2) 无插头：方案数不变。

2. 平地

    1) 有2个插头：当前决策的格子已经有路通过，因此下个状态是无插头。

    2) 有1个插头：当前决策的格子有路通过，但不确定向右还是向下，因此下个状态是1个插头。

    3) 有0个插头：当前决策的格子暂时没有路通过，要通一条路，因此下个状态是2个插头。

#include <cstdio>
#include <cstring>
#define rep(i,j,k) for(int i=j;i<k;++i)
#define FOR(i,j,k) for(int i=j;i<=k;++i)
const int N = 13, M = 1 << N;
typedef long long ll;
ll dp[N][N][M];
int n, m, mp[N][N];
int main() {
    int t;
    scanf("%d", &t);
    FOR(kase,1,t) {
        scanf("%d%d", &n, &m);
        FOR(i,1,n) FOR(j,1,m) scanf("%d", &mp[i][j]);
        memset(dp, 0, sizeof dp);
        dp[0][m][0] = 1;
        FOR(i,1,n) {
            rep(k,0,1<<m) dp[i][0][k << 1] = dp[i - 1][m][k];
            FOR(j,1,m) rep(k,0,1<<m+1) {
                int x = 1 << j - 1, y = 1 << j;
                if (!mp[i][j])
                    if (k & x || k & y) dp[i][j][k] = 0;
                    else dp[i][j][k] = dp[i][j - 1][k];
                else
                    if (((k & x) << 1) == (k & y)) // 2.1, 2.3, 插头状态相同时
                        dp[i][j][k] = dp[i][j - 1][k ^ x ^ y]; // 取相反的状态
                    else // 2.2
                        dp[i][j][k] = dp[i][j - 1][k] + dp[i][j - 1][k ^ x ^ y];
            }
        }
        printf("Case %d: There are %I64d ways to eat the trees.\n", kase, dp[n][m][0]);
    }
    return 0;
}

Eat the Trees

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3534    Accepted Submission(s): 1729

Problem Description

Most of us know that in the game called DotA(Defense of the Ancient), Pudge is a strong hero in the first period of the game. When the game goes to end however, Pudge is not a strong hero any more.
So Pudge’s teammates give him a new assignment—Eat the Trees!

The trees are in a rectangle N * M cells in size and each of the cells either has exactly one tree or has nothing at all. And what Pudge needs to do is to eat all trees that are in the cells.
There are several rules Pudge must follow:
I. Pudge must eat the trees by choosing a circuit and he then will eat all trees that are in the chosen circuit.
II. The cell that does not contain a tree is unreachable, e.g. each of the cells that is through the circuit which Pudge chooses must contain a tree and when the circuit is chosen, the trees which are in the cells on the circuit will disappear.
III. Pudge may choose one or more circuits to eat the trees.

Now Pudge has a question, how many ways are there to eat the trees?
At the picture below three samples are given for N = 6 and M = 3(gray square means no trees in the cell, and the bold black line means the chosen circuit(s))

 

Input

The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 10 cases.
For each case, the first line contains the integer numbers N and M, 1<=N, M<=11. Each of the next N lines contains M numbers (either 0 or 1) separated by a space. Number 0 means a cell which has no trees and number 1 means a cell that has exactly one tree.

 

Output

For each case, you should print the desired number of ways in one line. It is guaranteed, that it does not exceed 2 ⁶³ – 1. Use the format in the sample.

 

Sample Input

   
   
   
   

    
    
    
    2
6 3
1 1 1
1 0 1
1 1 1
1 1 1
1 0 1
1 1 1
2 4
1 1 1 1
1 1 1 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case 1: There are 3 ways to eat the trees.
Case 2: There are 2 ways to eat the trees.