POJ3254（状压DP）

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.

题意：给你一个农场，要往农场中种东西，有的地方不能种，相邻地方不能种，问总共有多少种种法。

题解：用状压dp，先用一个数组存储田地的初始状态。把不能种的地方储存为1。再用一个数组存储，同行没有相邻块的方案。再把这些同行没有相邻块的方案匹配到田地的第一行中。后面的行的方案数用状态转移方程：d[i][j] = (d[i - 1][k1] + d[i - 1][k2]+…) % MOD得出。最后输出的时候输出所有最后一行d[n][i]，i为可行方案编号。

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <map>#include <queue>#include <cmath>#include <vector>#include <functional>using namespace std;const int MOD = 100000000;const int maxn = 16;int n, m, a[maxn] = {0}, sta[1 << maxn] = {0}, num = 0, d[maxn][1 << maxn];void init(){    int sum = 1 << n;    for (int i = 0; i < sum; i++) {        if (i & (i << 1)) continue;        sta[num++] = i;    }}bool fit(int a, int b){    if (a & b) return false;    else return true;}void dp(){    for (int i = 0; i < num; i++) {        if (fit (sta[i], a[1])) d[1][i] = 1;    }    for (int i = 2; i <= m; i++) {        for (int j = 0; j < num; j++) {            if (fit (sta[j], a[i])) {                for (int k = 0; k < num; k++) {                    if (fit (sta[j], sta[k]) && fit (sta[k], a[i - 1])) {                        d[i][j] = (d[i][j] + d[i - 1][k]) % MOD;                    }                }            }        }    }}int main(){    #ifndef ONLINE_JUDGE    freopen ("in.txt", "r", stdin);    #endif // ONLINE_JUDGE    scanf ("%d%d", &m, &n);    for (int i = 1; i <= m; i++) {        for (int j = 1; j <= n; j++) {            int tmp;            scanf ("%d", &tmp);            if (tmp == 0) {                a[i] += 1 << (n - j);            }        }    }    init ();    dp ();    int cnt = 0;     for (int i = 0; i < num; i++) {        cnt = (cnt + d[m][i]) % MOD;     }     printf ("%d\n", cnt);    return 0;}