BZOJ 2669([cqoi2012]局部极小值-状态压缩+dp)

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有一个n行m列的整数矩阵，其中1到nm之间的每个整数恰好出现一次。如果一个格子比所有相邻格子（相邻是指有公共边或公共顶点）都小，我们说这个格子是局部极小值。
给出所有局部极小值的位置，你的任务是判断有多少个可能的矩阵。（1<=n<=4, 1<=m<=7）

状压dp+容斥容斥哪些X是此次DP的内容 dp[i][2^9] 表示该填i了 2^9表示哪些X被填了
首先dfs所有的X，看哪些X是这次DP要考虑的

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<ctime>#include<iomanip> #include<vector>#include<string>#include<queue>#include<stack>#include<map>#include<sstream>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (12345678)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase) printf("Case #%d: ",kcase);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} #define MAXN (100)int n,m;int p2[100];ll f[MAXN][1<<9];int g[MAXN],g2[MAXN],dir[8][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}};bool inside(int x,int y) {    return 1<=x&&1<=y&&x<=n&&y<=m;}int id[MAXN][MAXN];int idx(int i,int j) {    return (i-1)*m+j-1;}int x[MAXN],y[MAXN],siz=0;int b[50];int calc() {    if (siz>9) return 0;     memset(b,-1,sizeof(b));    For(i,n) For(j,m) id[i][j]=idx(i,j);    MEM(g) MEM(g2)    Rep(i,siz) {        int X=x[i],Y=y[i];        g2[i]=1<<idx(X,Y);        b[id[X][Y]]=i;        Rep(di,8) {            int a=X+dir[di][0],b=Y+dir[di][1];            if (inside(a,b)) {                int p=1<<idx(a,b);                g[i]|=p;            }        }    }    MEM(f)    f[0][0]=1;    int S=1<<siz;    int all=(1<<(n*m))-1;    Rep(i,n*m) {        Rep(st,S) if (f[i][st]) {            int now=all;            Rep(j,siz)                if (st&p2[j]) {                    now-=now&g2[j];                } else {                    now-=now&g[j];                }            int tot=0;            int tot1=0,tot2=0;            Rep(p,n*m) if (now&p2[p]) {                tot++;                if (-1 == b[p]) tot2++;                else tot1++;            }            int lef=n*m-i;            int choseX=siz-tot1;            int useBlank=i-choseX;            tot2-=useBlank;            tot2=min(tot2,lef-tot1);            if (tot2>0)                 upd(f[i+1][st],mul(tot2,f[i][st]));            Rep(p,n*m) {                if (now&p2[p]) {                    if (b[p]==-1) {                    } else {                        upd(f[i+1][st|p2[b[p]]],f[i][st]);                    }                }            }        }    }    return f[n*m][S-1];} bool ma[10][10];ll ans=0;void dfs(int dep,int l) {    if (dep==1) upd(ans,calc());    else ans=sub(ans,calc());    For(i,n) For(j,m) if (idx(i,j)>=l && !ma[i][j]){        bool fl=0;        Rep(di,8) {            int a=i+dir[di][0],b=j+dir[di][1];            if (!inside(a,b)) continue;            if (ma[a][b]) { fl=1; break;            }        }        if (!fl) {            ma[i][j]=1;            x[siz]=i,y[siz]=j;            ++siz;            dfs(-dep,idx(i,j)+1);            ma[i][j]=0;            --siz;        }    }}int main(){    p2[0]=1;    For(i,30) p2[i]=p2[i-1]*2;    int kcase=1;    while(cin>>n>>m) {        siz=0;        For(i,n) {            char s[10];            scanf("%s",s);            Rep(j,m) if (s[j]=='X') {                x[siz]=i; y[siz]=j+1;                ++siz; ma[i][j+1]=1;            } else ma[i][j+1]=0;        }        ans=0;        dfs(1,0);               cout<<ans<<endl;    }    return 0;}

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