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Corn Fields

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15285		Accepted: 8033

Description

Farmer John has purchased a lush new rectangular pasture composed of M by N (1 ≤ M ≤ 12; 1 ≤ N ≤ 12) square parcels. He wants to grow some yummy corn for the cows on a number of squares. Regrettably, some of the squares are infertile and can‘t be planted. Canny FJ knows that the cows dislike eating close to each other, so when choosing which squares to plant, he avoids choosing squares that are adjacent; no two chosen squares share an edge. He has not yet made the final choice as to which squares to plant.

Being a very open-minded man, Farmer John wants to consider all possible options for how to choose the squares for planting. He is so open-minded that he considers choosing no squares as a valid option! Please help Farmer John determine the number of ways he can choose the squares to plant.

Input

Line 1: Two space-separated integers: M and N
Lines 2..M+1: Line i+1 describes row i of the pasture with N space-separated integers indicating whether a square is fertile (1 for fertile, 0 for infertile)

Output

Line 1: One integer: the number of ways that FJ can choose the squares modulo 100,000,000.

Sample Input

2 3
1 1 1
0 1 0

Sample Output

9

Hint

Number the squares as follows:

1 2 3
  4  

There are four ways to plant only on one squares (1, 2, 3, or 4), three ways to plant on two squares (13, 14, or 34), 1 way to plant on three squares (134), and one way to plant on no squares. 4+3+1+1=9.

Source

USACO 2006 November Gold 題意：1個矩陣裏有很多格子，每個格子有兩種狀態，可以放牧和不可以放牧，可以放牧用1表示，否則用0表示，在這塊牧場放牛，要求兩個相鄰的方格不能同時放牛，即牛與牛不能相鄰。問有多少種放牛方案(一頭牛都不放也是一種方案) 題解：

【解析】根據題意，把每一行的狀態用二進制的數表示，0代表不在這塊放牛，1表示在這一塊放牛。首先很容易看到，每一行的狀態要符合牧場的硬件條件，即牛必須放在能放牧的方格上。這樣就能排除一些狀態。另外，牛與牛之間不能相鄰，這樣就要求每一行中不能存在兩個相鄰的1，這樣也能排除很多狀態。然後就是根據上一行的狀態轉移到當前行的狀態的問題了。必須符合不能有兩個1在同一列（兩只牛也不能豎著相鄰）的條件。這樣也能去掉一些狀態。然後，上一行的所有符合條件的狀態的總的方案數就是當前行該狀態的方案數。

【狀態表示】dp[state][i]：在狀態為state時，到第i行符合條件的可以放牛的方案數

【狀態轉移方程】dp[state][i] =Sigma dp[state‘][i-1] （state‘為符合條件的所有狀態）

【DP邊界條件】首行放牛的方案數dp[state][1] =1(state符合條件) OR 0 (state不符合條件)

以上解析 轉載需研讀

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<map>
 7 #include<queue>
 8 #include<stack>
 9 #include<vector>
10 #include<set>
11 #define ll __int64
12 #define mod 100000000
13 using namespace std;
14 int n,m;
15 int a[1003];
16 int b[1003];
17 int dp[20][1003];
18 int cnt=0;
19 bool check(int x)
20 {
21     if(x&(x/2)) return false;
22     else return true;
23 }
24 bool fun(int x,int k)
25 {
26     if(x&b[k]) return false;
27     else return true;
28 }
29 int main()
30 {
31     scanf("%d %d",&m,&n);
32     int exm;
33     for(int i=0;i<(1<<n);i++){
34         if(check(i))
35             a[++cnt]=i;
36     }
37     memset(b,0,sizeof(b));
38     for(int i=1;i<=m;i++)
39         for(int j=1;j<=n;j++){
40         scanf("%d",&exm);
41         if(exm==0)
42             b[i]+=(1<<(n-j));
43     }
44     for(int i=1;i<=cnt;i++){
45         if(fun(a[i],1))
46             dp[1][i]=1;
47     }
48     for(int i=2;i<=m;i++){
49         for(int k=1;k<=cnt;k++){
50         if(!fun(a[k],i)) continue;
51         for(int j=1;j<=cnt;j++){
52         if(!fun(a[j],i-1)) continue;
53         if(a[k]&a[j]) continue;
54         dp[i][k]=(dp[i][k]+dp[i-1][j])%mod;
55         }
56     }
57    }
58     int ans=0;
59     for(int i=1;i<=cnt;i++)
60         ans=(ans+dp[m][i])%mod;
61     printf("%d\n",ans);
62     return 0;
63 }

poj 3254 狀態壓縮