[DLX精确覆盖+打表] hdu 2518 Dominoes
Posted yfceshi
tags:
篇首语：本文由小常识网(cha138.com)小编为大家整理，主要介绍了[DLX精确覆盖+打表] hdu 2518 Dominoes相关的知识，希望对你有一定的参考价值。
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; }
以上是关于[DLX精确覆盖+打表] hdu 2518 Dominoes的主要内容，如果未能解决你的问题，请参考以下文章
POJ3740Easy Finding DLX（Dancing Links）精确覆盖问题