HDU-1693 Eat the Trees(插头DP)

Eat the Trees

题目链接
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 263 – 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.

偷懒只放个代码好了
关于插头DP最好去看看cdq的论文：基于连通性状态压缩的动态规划问题

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
#define inf 0x3f3f3f3f
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll MAXN = 1e6 + 7;
const ll MAXM = 1e3 + 7;
const ll MOD = 1e9 + 7;
const double eps = 1e-6;
int n, m, cur;
ll dp[2][1 << 12 + 2];
int G[15][15];
int main()
{
    int t;
    scanf("%d", &t);
    for (int Case = 1; Case <= t; Case++)
    {
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= m; j++)
                scanf("%d", &G[i][j]);
        if (n < m)
            swap(n, m);
        cur = 0;
        memset(dp[cur], 0, sizeof(dp[cur]));
        dp[cur][0] = 1;
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= m; j++)
            {
                cur ^= 1;
                memset(dp[cur], 0, sizeof(dp[cur]));
                for (int k = 0; k < (1 << m + 1); k++)
                {
                    if (!G[i][j]) //障碍物 无插头 直接转移
                    {
                        if (!(k & 1 || k & (1 << m))) //障碍物有左插头或者上插头肯定不行
                            dp[cur][k << 1] += dp[cur ^ 1][k];
                    }
                    else
                    {
                        if (j != 1)
                        {
                            if (k & (1 << m) && (k & 1)) /* 有上插头和左插头*/
                                dp[cur][(k ^ (1 << m | 1)) << 1] += dp[cur ^ 1][k];
                            else if (k & (1 << m)) /* 只有上插头 */
                            {
                                if (j != m) /* 右插头*/
                                    dp[cur][((k ^ (1 << m)) << 1) | 1] += dp[cur ^ 1][k];
                                if (i != n) /* 下插头 */
                                    dp[cur][((k ^ (1 << m)) | 1) << 1] += dp[cur ^ 1][k];
                            }
                            else if (k & 1) /* 只有左插头 */
                            {
                                if (j != m) /* 右插头 */
                                    dp[cur][(k ^ 1) << 1 | 1] += dp[cur ^ 1][k];
                                if (i != n) /*下插头 */
                                    dp[cur][k << 1] += dp[cur ^ 1][k];
                            }
                            else if (j != m && i != n) /* 没有插头只能右下插头 */
                                dp[cur][k << 1 | 3] += dp[cur ^ 1][k];
                        }
                        else if (!(k & 1)) /* 第一位必然不可能出现左插头的情况 */
                        {
                            if (k & (1 << m)) /* 有上插头 */
                            {
                                if (j != m) /* 右插头*/
                                    dp[cur][(k ^ (k & (1 << m))) << 1 | 1] += dp[cur ^ 1][k];
                                if (i != n) /* 下插头 */
                                    dp[cur][(k ^ (k & (1 << m))) << 1 | 2] += dp[cur ^ 1][k];
                            }
                            else if (i != n && j != m) /* 无插头 只能接右下*/
                                dp[cur][k << 1 | 3] += dp[cur ^ 1][k];
                        }
                    }
                }
            }
        }
        printf("Case %d: There are %lld ways to eat the trees.
", Case, dp[cur][0]);
    }
    return 0;
}