POJ P3254 Corn fields 【状压dp】

Corn Fields

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 16909		Accepted: 8939

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

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设f[i][s]表示第i行s状态下的方案数，则对于所有i - 1行与s不冲突的方案都可以转移

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define fo(i,x,y) for (int i = (x); i <= (y); i++)
#define Redge(u) for (int k = head[u]; k != -1; k = edge[k].next)
using namespace std;
const int maxn = 20,maxm = 1 << 12,INF = 1000000000,P =  100000000;

inline int read(){
	int out = 0,flag = 1;char c = getchar();
	while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57) {out = out * 10 + c - 48; c = getchar();}
	return out * flag;
}

int f[maxn][maxm],N,M,le[maxm];
bool ill[maxm];

int main()
{
	N = read();
	M = read();
	int maxv = (1 << M) - 1;
	for (int i = 1; i <= N; i++){
		for (int j = 1; j <= M; j++)
			le[i] = (le[i] << 1) + read();
	}
	for (int s = 0; s <= maxv; s++){
		bool sig = false; int t = s;
		while (t){
			if ((t & 1)&& sig) {ill[s] = true; break;}
			else if (t & 1) sig = true;
			else sig = false;
			t >>= 1;
		}
	}
	for (int s = 0; s <= maxv; s++){
		if (!ill[s] && (s | le[1]) == le[1])
			f[1][s] = 1;
	}
	for (int i = 2; i <= N; i++)
		for (int e = 0; e <= maxv; e++){
			if (ill[e] || (e | le[i]) != le[i]) continue;
			for(int s = 0; s <= maxv; s++){
				if (e & s) continue;
				f[i][e] = (f[i][e] + f[i - 1][s]) % P;
			}
		}
	int ans = 0;
	for (int i = 0; i <= maxv; i++) ans = (ans + f[N][i]) % P;
	cout<<ans<<endl;
	return 0;
}