TOJ 2850 Corn Fields 状压dp

Source: http://acm.tju.edu.cn/toj/showp.php?pid=2850

题意：n*m的土地上种东西，每个位置分为可以种和不能，种的方案要求不能相邻地种，问合法方案数。（写得有点乱）

分析：做过炮兵阵地，这题就是秒杀了。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int mo = 100000000;
int n, m, size, maxl;
int map[15][15];
int s[10000];
int dp[15][10000];
void dfs()
{
    size = 0;
    maxl = 1 << m;
    for (int i = 0; i < maxl; i++)
        if ((i & (i << 1)) == 0) s[size++] = i;
}
bool ok(int row, int id)
{
    int x = s[id], tmp;
    for (int i = m; i > 0; i--){
        tmp = x % 2;
        if (tmp == 1 && map[row][i] == 0) return false;
        x /= 2;
    }
    return true;
}
int main()
{
    while(scanf("%d %d", &n, &m) != EOF)
    {
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= m; j++)
                scanf("%d", &map[i][j]);
        memset(dp, 0, sizeof(dp));
        dp[0][0] = 1;
        dfs();
        for (int i = 1; i <= n; i++){
            for (int j = 0; j < size; j++){
                if (ok(i, j)){
                    for (int k = 0; k < size; k++)
                        if ((s[k] & s[j]) == 0) dp[i][j] = (dp[i][j] + dp[i-1][k]) % mo;
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < size; i++) ans = (ans + dp[n][i]) % mo;
        printf("%d\n", ans);
    }
    return 0;
}