题目链接
题意：给你一个n*m的矩阵，0代表不可以站人，1代表可以站人，并且相邻的格子之间不能站人（上下左右四个方向）。问有多少种站人的方法。
考虑每一行，便利每一种状态，在当前行的基础上，维护下一行的状态，复杂度 o(1<<m)^2*o(n)
左右相邻的用j&(j>>1)即可判断
上下相邻的用k&j可判断
/// .-~~~~~~~~~-._ _.-~~~~~~~~~-./// __.' ~. .~ `.__/// .'// \./ \\`./// .'// | \\`./// .'// .-~"""""""~~~~-._|_,-~~~~"""""""~-. \\`.///.'//.-" `-. | .-' "-.\\`./// .'//______.============-.. \ | / ..-============.______\\`./// .'______________________________\|/______________________________`.#pragma GCC optimize("Ofast")#pragma comment(linker, "/STACK:102400000,102400000")#pragma GCC target(sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx)#include<bits/stdc++.h>using namespace std;#define pi acos(-1)#define s_1(x) scanf("%d",&x)#define s_2(x,y) scanf("%d%d",&x,&y)#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)#define S_1(x) scan_d(x)#define S_2(x,y) scan_d(x),scan_d(y)#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)#define PI acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define fOR(n,x,i) for(int i=n;i>=x;i--)#define fOr(n,x,i) for(int i=n;i>x;i--)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db double//#define ll long long#define mp make_pair#define pb push_back#define sz size#define fi first#define se second#define pf printftypedef long long LL;typedef pair <int, int> ii;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=1e2+7;const int _=1e5+10;const double EPS=1e-8;const double eps=1e-8;const LL mod=100000000;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}template <class T>inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}if(IsN) num=-num;return true;}void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}void print(LL a){ Out(a),puts("");}//cerr << "run time is " << clock() << endl;int n,m;int a[_];int dp[15][1<<13];void solve() {W(s_2(n,m)!=EOF) {me(dp,0);me(a,0);dp[0][0]=1;FOR(1,n,i)FOr(0,m,j) {int x;s_1(x);a[i]|=(x<<j);}int sta=1<<m;FOR(1,n,i) {for(int j=0;j<sta;j++) {if((j&a[i])!=j) continue;if(j&(j>>1)) continue;for(int k=0;k<sta;k++) {if(k&j) continue;dp[i][j]=(dp[i][j]+dp[i-1][k])%mod;}}}int ans=0;for(int i=0;i<sta;i++)ans=(ans+dp[n][i])%mod;print(ans);}}int main() {//freopen( "1.in" , "r" , stdin );//freopen( "1.out" , "w" , stdout );int t=1;//init();//s_1(t);for(int cas=1;cas<=t;cas++) {//printf("Case %d:\n",cas);solve();}}
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