HDU 1693 新的恐懼，插頭Dp

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 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.

 

題目大意就是求完全覆蓋的合法方案數。

插頭Dp。感受到了博弈論時候的恐懼QAQ。

那麼我的f[i][j][S]表示處理完(i,j)這個格子後插頭的集合QWQ。

那麼f[i][0][S<<1]=f[i-1][m][S];因爲枚舉處理完上一行的狀態來更新這一行的狀態。

f[i][j][k]=f[i][j-1][k^(1<<(j-1))^(1<<j)]考慮每個方塊有一個或兩個的都能考慮進去(除去單個豎直插頭)

f[i][j][k]=f[i][j-1][k]當這個方塊只有一個豎直插頭的時候可以由上一個方塊轉移。

上面討論的都是非障礙，下面討論障礙。

如果當前障礙(i,j)上沒有插頭則繼承上一個即f[i][j][S]=f[i][j-1][S];

然後答案就是f[n][m][0]即操作完(n,m)後剩餘插頭狀態爲0(因爲輪廓線下移。)

code：

#define ll long long


#include <stdio.h>
#include <cstring>
#include <iostream>


template<typename _t>
inline _t read(){
    _t x=0,f=1;
    char ch=getchar();
    for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-f;
    for(;isdigit(ch);ch=getchar())x=x*10+(ch^48);
    return x*f;
}

int n,m,a[13][13],Tcase,num;
ll f[13][13][1<<13];

void Dp(){
    memset(f,0,sizeof f);
    f[0][m][0]=1;
    int full = 1<<(m+1);
    for(int i=1;i<=n;i++){
        for(int j=0;j<full>>1;j++)
            f[i][0][j<<1]=f[i-1][m][j];
        for(int j=1;j<=m;j++)
            for(int k=0;k<full;k++){
                int x = 1<<(j-1);
                int y = 1<<(j);
                if(a[i][j]){
                    f[i][j][k]+=f[i][j-1][k^x^y];
                    if((k&x)&&(k&y))continue;
                    if(!(k&x)&&!(k&y))continue;
                    f[i][j][k]+=f[i][j-1][k];
                }
                else{
                    if(!(k&x)&&!(k&y))f[i][j][k]+=f[i][j-1][k];
                    else f[i][j][k]=0;
                }
            }
    }
}

int main(){
    Tcase=read<int>();
    while(Tcase--){
        n=read<int>(),m=read<int>();
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                a[i][j]=read<int>();
        Dp();
        printf("Case %d: There are %lld ways to eat the trees.\n",++num,f[n][m][0]);
    }
}