【USACO 2006 November Gold】Corn Fields

【题目链接】
          点击打开链接
【算法】
         状压DP
【代码】
    #include<bits/stdc++.h>
using namespace std;
#define MAXN 12
#define MOD 100000000

int M,N,i,j,k,ans,state;
int ST[MAXN+1][(1<<12)+1],f[MAXN+1][(1<<12)+1],cnt[MAXN+1],mat[MAXN+1][MAXN+1];

template <typename T> inline void read(T &x) {
int f = 1; x = 0;
char c = getchar();
for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
x *= f;
}
template <typename T> inline void write(T x) {
if (x < 0) { putchar('-'); x = -x; }
if (x > 9) write(x/10);
putchar(x%10+'0');
}
template <typename T> inline void writeln(T x) {
write(x);
puts("");
}

inline void dfs(int dep) {
if (dep > N) ST[i][++cnt[i]] = state;
else {
dfs(dep+1);
if (mat[i][dep]) {
if ((dep == 1) || (!(state & (1 << (N - dep + 1))))) {
state |= (1 << (N - dep));
dfs(dep+1);
state &= (~(1 << (N - dep)));
}
}
}
}

int main() {

read(M); read(N);

for (i = 1; i <= M; i++) {
for (j = 1; j <= N; j++) {
read(mat[i][j]);
}
}

for (i = 1; i <= M; i++) dfs(1);

for (i = 1; i <= cnt[1]; i++) f[1][i] = 1;
for (i = 2; i <= M; i++) {
for (j = 1; j <= cnt[i]; j++) {
for (k = 1; k <= cnt[i-1]; k++) {
if (!(ST[i][j] & ST[i-1][k]))
f[i][j] = (f[i][j] + f[i-1][k]) % MOD;
}
}
}

for (i = 1; i <= cnt[M]; i++) ans = (ans + f[M][i]) % MOD;
writeln(ans);

return 0;

}