标签:sizeof push this 问题 dlx tor fine 算法 cst
DLX算法求解精确覆盖问题模板。赛场上可以参见白书。
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
const int sub[10][10]={
{0,0,0,0,0,0,0,0,0,0},
{0,1,1,1,2,2,2,3,3,3},
{0,1,1,1,2,2,2,3,3,3},
{0,1,1,1,2,2,2,3,3,3},
{0,4,4,4,5,5,5,6,6,6},
{0,4,4,4,5,5,5,6,6,6},
{0,4,4,4,5,5,5,6,6,6},
{0,7,7,7,8,8,8,9,9,9},
{0,7,7,7,8,8,8,9,9,9},
{0,7,7,7,8,8,8,9,9,9}
};
int hlb[12],hub[12],llb[12],lub[12];
const int maxn=4*9*9+5;
const int maxr=9*9*9+5;
const int maxnode=9*9*9*4+maxn+5;
//ÐбàºÅ´Ó1¿ªÊ¼£¬ÁбàºÅΪ1~n£¬½áµã0ÊDZíÍ·½áµã£»½áµã1~nÊǸ÷Áж¥²¿µÄÐéÄâ½áµã
struct DLX{
int n,sz;//ÁÐÊý£¬½áµã×ÜÊý
int S[maxn];//¸÷ÁнáµãÊý
int row[maxnode],col[maxnode];//¸÷½áµãËùÔÚµÄÐÐÁбàºÅ
int L[maxnode],R[maxnode],U[maxnode],D[maxnode];
int ansd,ans[maxr];//½â
void init(int n){//nÊÇÁÐÊý
this->n=n;
for(int i=0;i<=n;++i){
U[i]=i;
D[i]=i;
L[i]=i-1;
R[i]=i+1;
}
R[n]=0; L[0]=n;
sz=n+1;
memset(S,0,sizeof(S));
}
void addRow(int r,vector<int> columns){
int first=sz;
for(int i=0;i<columns.size();++i){
int c=columns[i];
L[sz]=sz-1;
R[sz]=sz+1;
D[sz]=c;
U[sz]=U[c];
D[U[c]]=sz;
U[c]=sz;
row[sz]=r;
col[sz]=c;
++S[c];
++sz;
}
R[sz-1]=first;
L[first]=sz-1;
}
//˳×ÅÁ´±íA£¬±éÀú³ýsÍâµÄÆäËûÔªËØ
#define FOR(i,A,s) for(int i=A[s];i!=s;i=A[i])
void remove(int c){
L[R[c]]=L[c];
R[L[c]]=R[c];
FOR(i,D,c){
FOR(j,R,i){
U[D[j]]=U[j];
D[U[j]]=D[j];
--S[col[j]];
}
}
}
void restore(int c){
FOR(i,U,c){
FOR(j,L,i){
++S[col[j]];
U[D[j]]=j;
D[U[j]]=j;
}
}
L[R[c]]=c;
R[L[c]]=c;
}
bool dfs(int d){
// printf("%d",d);
if(R[0]==0){//ÕÒµ½½â
ansd=d;//¼Ç¼½âµÄ³¤¶È
return 1;
}
//ÕÒ½áµãÊý×îСµÄÁÐc
int c=R[0];//µÚÒ»¸öδɾ³ýµÄÁÐ
FOR(i,R,0){
if(S[i]<S[c]){
c=i;
}
}
remove(c);//ɾ³ýµÚcÁÐ
FOR(i,D,c){//ÓýáµãiËùÔÚÐи²¸ÇµÚcÁÐ
ans[d]=row[i];
FOR(j,R,i){
remove(col[j]);//ɾ³ý½áµãiËùÔÚÐÐÄܸ²¸ÇµÄËùÓÐÆäËûÁÐ
}
if(dfs(d+1)){
return 1;
}
FOR(j,L,i){
restore(col[j]);//»Ö¸´½áµãiËùÔÚÐÐÄܸ²¸ÇµÄÆäËûËùÓÐÁÐ
}
}
restore(c);//»Ö¸´µÚcÁÐ
return 0;
}
bool solve(vector<int>& v){
v.clear();
if(!dfs(0)){
return 0;
}
for(int i=0;i<ansd;++i){
v.push_back(ans[i]);
}
return 1;
}
}dlx;
char s[12][12];
int a[12][12];
int mah[1005],mal[1005],mav[1005];
int encode(int a,int b,int c){
return a*81+b*9+c+1;
}
void decode(int code,int &a,int &b,int &c){
--code;
c=code%9;code/=9;
b=code%9;code/=9;
a=code;
}
int main(){
int zu;
// freopen("poj2676.in","r",stdin);
// freopen("poj2676.out","w",stdout);
memset(hlb,0x7f,sizeof(hlb));
memset(llb,0x7f,sizeof(llb));
for(int i=1;i<=9;++i){
for(int j=1;j<=9;++j){
hlb[sub[i][j]]=min(hlb[sub[i][j]],i);
hub[sub[i][j]]=max(hub[sub[i][j]],i);
llb[sub[i][j]]=min(llb[sub[i][j]],j);
lub[sub[i][j]]=max(lub[sub[i][j]],j);
}
}
scanf("%d",&zu);
for(;zu;--zu){
for(int i=1;i<=9;++i){
scanf("%s",s[i]+1);
}
int hang=0,lie=0;
for(int i=1;i<=9;++i){
for(int j=1;j<=9;++j){
a[i][j]=s[i][j]-‘0‘;
}
}
dlx.init(9*9*4);
for(int i=1;i<=9;++i){
for(int j=1;j<=9;++j){
for(int k=1;k<=9;++k){
if(!a[i][j] || a[i][j]==k){
vector<int> columns;
columns.push_back(encode(0,i-1,j-1));
columns.push_back(encode(1,i-1,k-1));
columns.push_back(encode(2,j-1,k-1));
columns.push_back(encode(3,sub[i][j]-1,k-1));
dlx.addRow(encode(i-1,j-1,k-1),columns);
}
}
}
}
vector<int> ans;
dlx.solve(ans);
for(int i=0;i<ans.size();++i){
int r,c,v;
decode(ans[i],r,c,v);
a[r+1][c+1]=v+1;
}
for(int i=1;i<=9;++i){
for(int j=1;j<9;++j){
printf("%d",a[i][j]);
}
printf("%d\n",a[i][9]);
}
}
return 0;
}
标签:sizeof push this 问题 dlx tor fine 算法 cst
原文地址:http://www.cnblogs.com/autsky-jadek/p/7752121.html