poj 3254 状态压缩dp

Corn Fields

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7878		Accepted: 4206

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

题意：1为草地 0为空地，草地可以放牛，但相邻的草地不可以都放牛，问有多少种放牛的方法。

思路：经典状态压缩问题，预处理单行的所有不相邻的可行情况，存到ans数组中，然后在dp的时候判断上下行是否有相邻格子，用ans的下标去代替状态的话，状态数最多也就几百个，数组就不用开到2^12，大大节约了空间，还可以再用滚动数组继续优化。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int vis[13][13];
int dp[2][500];
int ans[500];
int t=0;
int n,m;
void init(int n)
{
for(int i=0; i<(1<<n); i++){
int c=i;
while(!((c&1)&&(c&2))){
if(c<3){ans[t++]=i;break;}
c>>=1;
}
}
}
int pd(int a,int i)
{
for(int j=1; j<=m; j++)
if(vis[i][j]==0&&(a&(1<<(j-1))))return 0;
return 1;
}
int pds(int a,int b)
{
while(a>0){
if((a&1)&(b&1))return 0;
a>>=1;
b>>=1;
}
return 1;
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF){
init(m);
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
scanf("%d",&vis[i][j]);
for(int i=0;i<t;i++)
if(pd(ans[i],1))
dp[1][i]=1;
for(int i=2;i<=n;i++){
memset(dp[i&1],0,sizeof(dp[i&1]));
for(int j=0;j<t;j++)
for(int k=0;k<t;k++)
if(pd(ans[j],i)&&pds(ans[j],ans[k]))
dp[i&1][j]+=dp[i&1^1][k];
}
int sum=0;
for(int i=0;i<t;i++){
sum+=dp[n&1][i];
sum%=100000000;
}
printf("%d\n",sum);
}
return 0;
}