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[DLX精确覆盖+打表] hdu 2518 Dominoes

题意:

就是给12种图形,旋转,翻折。有多少种方法构成n*m=60的矩形

思路:

裸的精确覆盖。就是建图麻烦

个人太挫,直接手写每一个图形的各种形态

须要注意的是最后的答案须要除以4

代码:

#include"stdio.h"
#include"algorithm"
#include"string.h"
#include"iostream"
#include"queue"
#include"map"
#include"vector"
#include"string"
using namespace std;
/*int mp[63][5][5]=
{
    {
        //1.1
        {1,0,0},
        {1,0,0},
        {1,1,1},
    },
    {
        //1.2
        {1,1,1},
        {0,0,1},
        {0,0,1},
    },
    {
        //1.3
        {0,0,1},
        {0,0,1},
        {1,1,1},
    },
    {
        //1.4
        {1,1,1},
        {1,0,0},
        {1,0,0},
    },
    {
        //2.5
        {1,1,1,1,1},
    },
    {
        //2.6
        {1},
        {1},
        {1},
        {1},
        {1},
    },
    {
        //3.7
        {0,1,0},
        {1,1,1},
        {0,1,0},
    },
    {
        //4.8
        {1,1,1},
        {1,0,1},
    },
    {
        //4.9
        {1,0,1},
        {1,1,1},
    },
    {
        //4.10
        {1,1},
        {1,0},
        {1,1},
    },
    {
        //4.11
        {1,1},
        {0,1},
        {1,1},
    },
    {
        //5.12
        {1,1,1,1},
        {1,0,0,0},
    },
    {
        //5.13
        {1,0},
        {1,0},
        {1,0},
        {1,1},
    },
    {
        //5.14
        {0,0,0,1},
        {1,1,1,1},
    },
    {
        //5.15
        {1,1},
        {0,1},
        {0,1},
        {0,1},
    },
    {
        //5.16
        {1,0,0,0},
        {1,1,1,1},
    },
    {
        //5.17
        {0,1},
        {0,1},
        {0,1},
        {1,1},
    },
    {
        //5.18
        {1,1,1,1},
        {0,0,0,1},
    },
    {
        //5.19
        {1,1},
        {1,0},
        {1,0},
        {1,0},
    },
    {
        //6.20
        {1,0,0},
        {1,1,0},
        {0,1,1},
    },
    {
        //6.21
        {0,0,1},
        {0,1,1},
        {1,1,0},
    },
    {
        //6.22
        {1,1,0},
        {0,1,1},
        {0,0,1},
    },
    {
        //6.23
        {0,1,1},
        {1,1,0},
        {1,0,0},
    },
    {
        //7.24
        {1,1,1,1},
        {0,1,0,0},
    },
    {
        //7.25
        {1,0},
        {1,0},
        {1,1},
        {1,0},
    },
    {
        //7.26
        {0,0,1,0},
        {1,1,1,1},
    },
    {
        //7.27
        {0,1},
        {1,1},
        {0,1},
        {0,1},
    },
    {
        //7.28
        {0,1,0,0},
        {1,1,1,1},
    },
    {
        //7.29
        {0,1},
        {0,1},
        {1,1},
        {0,1},
    },
    {
        //7.30
        {1,1,1,1},
        {0,0,1,0},
    },
    {
        //7.31
        {1,0},
        {1,1},
        {1,0},
        {1,0},
    },
    {
        //8.32
        {0,0,1},
        {1,1,1},
        {1,0,0},
    },
    {
        //8.33
        {1,1,0},
        {0,1,0},
        {0,1,1},
    },
    {
        //8.34
        {1,0,0},
        {1,1,1},
        {0,0,1},
    },
    {
        //8.35
        {0,1,1},
        {0,1,0},
        {1,1,0},
    },
    {
        //9.36
        {0,1,0},
        {0,1,1},
        {1,1,0},
    },
    {
        //9.37
        {0,1,0},
        {1,1,1},
        {0,0,1},
    },
    {
        //9.38
        {0,1,1},
        {1,1,0},
        {0,1,0},
    },
    {
        //9.39
        {1,0,0},
        {1,1,1},
        {0,1,0},
    },
    {
        //9.40
        {1,1,0},
        {0,1,1},
        {0,1,0},
    },
    {
        //9.41
        {0,1,0},
        {1,1,1},
        {1,0,0},
    },
    {
        //9.42
        {0,1,0},
        {1,1,0},
        {0,1,1},
    },
    {
        //9.43
        {0,0,1},
        {1,1,1},
        {0,1,0},
    },
    {
        //10.44
        {0,1,0},
        {0,1,0},
        {1,1,1},
    },
    {
        //10.45
        {1,1,1},
        {0,1,0},
        {0,1,0},
    },
    {
        //10.46
        {0,0,1},
        {1,1,1},
        {0,0,1},
    },
    {
        //10.47
        {1,0,0},
        {1,1,1},
        {1,0,0},
    },
    {
        //11.48
        {0,1,1},
        {1,1,1},
    },
    {
        //11.49
        {1,1},
        {1,1},
        {0,1},
    },
    {
        //11.50
        {1,1,1},
        {1,1,0},
    },
    {
        //11.51
        {1,0},
        {1,1},
        {1,1},
    },
    {
        //11.52
        {1,1,1},
        {0,1,1},
    },
    {
        //11.53
        {1,1},
        {1,1},
        {1,0},
    },
    {
        //11.54
        {1,1,0},
        {1,1,1},
    },
    {
        //11.55
        {0,1},
        {1,1},
        {1,1},
    },
    {
        //12.56
        {0,1,1,1},
        {1,1,0,0},
    },
    {
        //12.57
        {1,0},
        {1,0},
        {1,1},
        {0,1},
    },
    {
        //12.58
        {0,0,1,1},
        {1,1,1,0},
    },
    {
        //12.59
        {1,0},
        {1,1},
        {0,1},
        {0,1},
    },
    {
        //12.60
        {1,1,1,0},
        {0,0,1,1},
    },
    {
        //12.61
        {0,1},
        {1,1},
        {1,0},
        {1,0},
    },
    {
        //12.62
        {1,1,0,0},
        {0,1,1,1},
    },
    {
        //12.63
        {0,1},
        {0,1},
        {1,1},
        {1,0},
    },
};
//a代表每一个的行,b代表每一个的列。c代表每一个属于哪种
int a[]= {3,3,3,3,1,5,3,2,2,3,3,2,4,2,4,2,4,2,4,3,3,3,3,2,4,2,4,2,4,2,4,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,2,3,2,3,2,3,2,3,2,4,2,4,2,4,2,4};
int b[]= {3,3,3,3,5,1,3,3,3,2,2,4,2,4,2,4,2,4,2,3,3,3,3,4,2,4,2,4,2,4,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,2,3,2,3,2,3,2,4,2,4,2,4,2,4,2};
int c[]= {1,1,1,1,2,2,3,4,4,4,4,5,5,5,5,5,5,5,5,6,6,6,6,7,7,7,7,7,7,7,7,8,8,8,8,9,9,9,9,9,9,9,9,10,10,10,10,11,11,11,11,11,11,11,11,12,12,12,12,12,12,12,12};
#define N 63*66*(60+66+14)
#define M 63*66
int ooo,haha;
struct DLX
{
    int n,m,C;
    int U[N],D[N],L[N],R[N],Row[N],Col[N];
    int H[M],S[M],cnt,ans[M];
    void init(int _n,int _m)
    {
        n=_n;
        m=_m;
        for(int i=0; i<=m; i++)
        {
            U[i]=D[i]=i;
            L[i]=(i==0?m:i-1);
            R[i]=(i==m?

0:i+1); S[i]=0; } C=m; for(int i=1; i<=n; i++) H[i]=-1; } void link(int x,int y) { C++; Row[C]=x; Col[C]=y; S[y]++; U[C]=U[y]; D[C]=y; D[U[y]]=C; U[y]=C; if(H[x]==-1) H[x]=L[C]=R[C]=C; else { L[C]=L[H[x]]; R[C]=H[x]; R[L[H[x]]]=C; L[H[x]]=C; } } void del(int x) { R[L[x]]=R[x]; L[R[x]]=L[x]; for(int i=D[x]; i!=x; i=D[i]) { for(int j=R[i]; j!=i; j=R[j]) { U[D[j]]=U[j]; D[U[j]]=D[j]; S[Col[j]]--; } } } void rec(int x) { for(int i=U[x]; i!=x; i=U[i]) { for(int j=L[i]; j!=i; j=L[j]) { U[D[j]]=j; D[U[j]]=j; S[Col[j]]++; } } R[L[x]]=x; L[R[x]]=x; } void dance(int x) { if(R[0]==0 || R[0]>ooo) { haha++; //cnt=x; return ; } int now=R[0]; for(int i=R[0]; i!=0 && i<=ooo; i=R[i]) { if(S[i]<S[now]) now=i; } del(now); for(int i=D[now]; i!=now; i=D[i]) { //ans[x]=Row[i]; for(int j=R[i]; j!=i; j=R[j]) del(Col[j]); dance(x+1); for(int j=L[i]; j!=i; j=L[j]) rec(Col[j]); } rec(now); return ; } } dlx; int main() { int n,m; while(scanf("%d%d",&n,&m)!=-1) { int cnt=0; ooo=60; dlx.init(63*60*5,60+12); for(int i=0; i<63; i++) { for(int xx=1; xx+a[i]<=n+1; xx++) { for(int yy=1; yy+b[i]<=m+1; yy++) { cnt++; // if(c[i]<5) printf("%d:",c[i]); for(int x=0; x<a[i]; x++) { for(int y=0; y<b[i]; y++) { if(mp[i][x][y]==1) { int tep=(xx+x-1)*m+(yy+y); // if(c[i]<5)printf("%d ",tep); dlx.link(cnt,tep); } } } //if(c[i]<5) puts(""); dlx.link(cnt,60+c[i]); } } } haha=0; dlx.dance(0); printf("%d\n",haha); } return 0; }*/ int ans[]={0,0,0,2,368,1010,2339}; int main() { int n,m; while(scanf("%d%d",&n,&m)!=-1) { if(n>m) swap(n,m); printf("%d\n",ans[n]); } return 0; }




[DLX精确覆盖+打表] hdu 2518 Dominoes