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 **************************************************************************************************** 状态压缩（用二进制）和炮兵阵相似 **************************************************************************************************** 1 /*
2 状态压缩用二进制，
3 先存储每一行的可行状态，
4 再判断；
5 */
6 #include<iostream>
7 #include<string>
8 #include<cstring>
9 #include<cstdio>
10 #include<cmath>
11 #include<algorithm>
12 using namespace std;
13 const int maxn=1<<12;
14 const int MOD=100000000;
15 int s[maxn],state[maxn];
16 int i,j,k,n,m,ht;
17 int a[maxn],b[maxn];//a[i]表示i个状态的状态数量,b[]中间数组
18 int num;
19 int main()
20 {
21 num=0;
22 for(i=0;i<=maxn;i++)
23 if((i&(i<<1))==0)//存储一行中的可利用状态
24 s[++num]=i;
25 cin>>n>>m;
26 for(i=1;i<=n;i++)
27 {
28 state[i]=0;
29 for(j=1;j<=m;j++)
30 {
31 cin>>k;
32 state[i]=(state[i]<<1)+1-k;//0表示可用，1表示不可用
33 }
34 }
35 k=1<<m;
36 for(j=1;s[j]<k;j++)
37 if(s[j]&state[1])
38 a[j]=0;
39 else
40 a[j]=1;
41 for(i=2;i<=n;i++)
42 {
43 for(j=1;s[j]<k;j++)
44 {
45 b[j]=0;
46 if(!(s[j]&state[i]))
47 {
48 for(ht=1;s[ht]<k;ht++)
49 if(!(s[j]&s[ht]))//本行与上一行的状态相契合
50 {
51 b[j]=(b[j]+a[ht])%MOD;
52 }
53 }
54 }
55
56 for(j=1;s[j]<k;j++)
57 a[j]=b[j];
58 }
59 int ans=0;
60 for(j=1;s[j]<k;j++)
61 ans=(ans+a[j])%MOD;
62 cout<<ans<<endl;
63 return 0;
64
65 }
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坚持！！！！！！
转载于:https://www.cnblogs.com/sdau--codeants/p/3307835.html