POJ 3074,3076,2676 数独 Dancing Links舞蹈链

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POJ 3074 Sudoku

思路：数独正好是精确覆盖问题的最好例子，这个也算是舞蹈链的模板了吧。

这个和网络流一样，难的是建模。如果能想好怎么建立矩阵的话，那一切都容易了。

这题怎么建矩阵，可以看论文：http://www.cnblogs.com/steady/archive/2011/03/15/1984791.html

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<map>#include<queue>#include<set>#include<cmath>#include<bitset>#define mem(a,b) memset(a,b,sizeof(a))#define lson i<<1,l,mid#define rson i<<1|1,mid+1,r#define llson j<<1,l,mid#define rrson j<<1|1,mid+1,r#define INF 0x7fffffff#define maxn 1000005typedef long long ll;typedef unsigned long long ull;using namespace std;int head,sz;int U[maxn],D[maxn],L[maxn],R[maxn];//上下左右链表指针int H[maxn],ROW[maxn],C[maxn],S[maxn],O[maxn];void remove(int c){    L[R[c]]=L[c];    R[L[c]]=R[c];    for(int i=D[c]; i!=c; i=D[i])        for(int j=R[i]; j!=i; j=R[j])        {            U[D[j]]=U[j];            D[U[j]]=D[j];            --S[C[j]];        }}void resume(int c){    for(int i=U[c]; i!=c; i=U[i])    {        for(int j=L[i]; j!=i; j=L[j])        {            ++S[C[j]];            U[D[j]]=j;            D[U[j]]=j;        }    }    L[R[c]]=c;    R[L[c]]=c;}void init(int m)//m是列{    head=0;//头指针为0    for(int i=0; i<=m; i++)    {        U[i]=i;        D[i]=i;//建立双向十字链表        L[i]=i-1;        R[i]=i+1;        S[i]=0;    }    R[m]=0;    L[0]=m;    S[0]=INF+1;    sz=m+1;    memset(H,0,sizeof(H));}void insert(int i, int j){    if(H[i])    {        L[sz] = L[H[i]];        R[sz] = H[i];        L[R[sz]] = sz;        R[L[sz]] = sz;    }    else    {        L[sz] = sz;        R[sz] = sz;        H[i] = sz;    }    U[sz] = U[j];    D[sz] = j;    U[D[sz]] = sz;    D[U[sz]] = sz;    C[sz] = j;    ROW[sz] = i;    ++S[j];    ++sz;}bool dfs(int k,int len){    if(R[head]==head)    {        sort(O,O+len*len);        int p=0;        for(int i=0; i<len; i++)            for(int j=0; j<len; j++)            {                int num=O[p++];                num=num-(i*len+j)*len;                printf("%d",num);            }        puts("");        return  true;    }    int s=INF,c;    for (int t=R[head]; t!=head; t=R[t])        if (S[t]<s) s=S[t],c=t;    remove(c);    for(int i=D[c]; i!=c; i=D[i])    {        O[k]=ROW[i];        for(int j=R[i]; j!=i; j=R[j])            remove(C[j]);        if(dfs(k+1,len)) return  true;        for(int j=L[i]; j!=i; j=L[j])            resume(C[j]);    }    resume(c);    return  false;}void calc(int i,int j,int k,int len){    int r=(i*len+j-1)*len+k;    int base=sqrt(len);    //第i行有数字k    insert(r,i*len+k);    //第j列有数字k    insert(r,len*len+(j-1)*len+k);    //第k块有数字k    int block=(j-1)/base*base+i/base;    insert(r,len*len*2+block*len+k);    //第i行j列有一个数字(限制一个出格子只填一个数)    insert(r,len*len*3+i*len+j);}void build(char *s,int len)//len表示是几宫数独{    int i,j,k,p=0;    init(len*len*4);    for(i=0; i<len; i++)        for(j=1; j<=len; j++,p++)        {            if(s[p]=='.')                for(k=1; k<=len; k++)                    calc(i,j,k,len);            else calc(i,j,s[p]-'0',len);        }}int main(){    //freopen("1.txt","r",stdin);    char s[84];    while(scanf("%s",s)&&strcmp(s,"end"))    {        build(s,9);        dfs(0,9);//从根开始搜    }    return 0;}

poj 3076 Sudoku

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<map>#include<queue>#include<set>#include<cmath>#include<bitset>#define mem(a,b) memset(a,b,sizeof(a))#define lson i<<1,l,mid#define rson i<<1|1,mid+1,r#define llson j<<1,l,mid#define rrson j<<1|1,mid+1,r#define INF 0x7fffffff#define maxn 1000005typedef long long ll;typedef unsigned long long ull;using namespace std;int head,sz;int U[maxn],D[maxn],L[maxn],R[maxn];//上下左右链表指针int H[maxn],ROW[maxn],C[maxn],S[maxn],O[maxn];void remove(int c){    L[R[c]]=L[c];    R[L[c]]=R[c];    for(int i=D[c]; i!=c; i=D[i])        for(int j=R[i]; j!=i; j=R[j])        {            U[D[j]]=U[j];            D[U[j]]=D[j];            --S[C[j]];        }}void resume(int c){    for(int i=U[c]; i!=c; i=U[i])    {        for(int j=L[i]; j!=i; j=L[j])        {            ++S[C[j]];            U[D[j]]=j;            D[U[j]]=j;        }    }    L[R[c]]=c;    R[L[c]]=c;}void init(int m)//m是列{    head=0;//头指针为0    for(int i=0; i<=m; i++)    {        U[i]=i;        D[i]=i;//建立双向十字链表        L[i]=i-1;        R[i]=i+1;        S[i]=0;    }    R[m]=0;    L[0]=m;    S[0]=INF+1;    sz=m+1;    memset(H,0,sizeof(H));}void insert(int i, int j){    if(H[i])    {        L[sz] = L[H[i]];        R[sz] = H[i];        L[R[sz]] = sz;        R[L[sz]] = sz;    }    else    {        L[sz] = sz;        R[sz] = sz;        H[i] = sz;    }    U[sz] = U[j];    D[sz] = j;    U[D[sz]] = sz;    D[U[sz]] = sz;    C[sz] = j;    ROW[sz] = i;    ++S[j];    ++sz;}bool dfs(int k,int len){    if(R[head]==head)    {        sort(O,O+len*len);        int p=0;        for(int i=0; i<len; i++)        {            for(int j=0; j<len; j++)            {                int num=O[p++];                num=num-(i*len+j)*len;                printf("%c",num+'A'-1);            }            puts("");        }        puts("");        return  true;    }    int s=INF,c;    for (int t=R[head]; t!=head; t=R[t])        if (S[t]<s) s=S[t],c=t;    remove(c);    for(int i=D[c]; i!=c; i=D[i])    {        O[k]=ROW[i];        for(int j=R[i]; j!=i; j=R[j])            remove(C[j]);        if(dfs(k+1,len))            return  true;        for(int j=L[i]; j!=i; j=L[j])            resume(C[j]);    }    resume(c);    return  false;}void calc(int i,int j,int k,int len){    int r=(i*len+j-1)*len+k;    int base=sqrt(len);    //第i行有数字k    insert(r,i*len+k);    //第j列有数字k    insert(r,len*len+(j-1)*len+k);    //第k块有数字k    int block=(j-1)/base*base+i/base;    insert(r,len*len*2+block*len+k);    //第i行j列有一个数字(限制一个出格子只填一个数)    insert(r,len*len*3+i*len+j);}void build(char s[][17],int len)//len表示是几宫数独{    int i,j,k;    init(len*len*4);    for(i=0; i<len; i++)        for(j=1; j<=len; j++)        {            if(s[i][j-1]=='-')                for(k=1; k<=len; k++)                    calc(i,j,k,len);            else calc(i,j,s[i][j-1]-'A'+1,len);        }}int main(){    //freopen("1.txt","r",stdin);    char s[17][17];    while(~scanf("%s",s[0]))    {        for(int i=1; i<16; i++)            scanf("%s",s[i]);        build(s,16);        dfs(0,16);//从根开始搜    }    return 0;}

POJ 2676 Sudoku

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<map>#include<queue>#include<set>#include<cmath>#include<bitset>#define mem(a,b) memset(a,b,sizeof(a))#define lson i<<1,l,mid#define rson i<<1|1,mid+1,r#define llson j<<1,l,mid#define rrson j<<1|1,mid+1,r#define INF 0x7fffffff#define maxn 1000005typedef long long ll;typedef unsigned long long ull;using namespace std;int head,sz;int U[maxn],D[maxn],L[maxn],R[maxn];//上下左右链表指针int H[maxn],ROW[maxn],C[maxn],S[maxn],O[maxn];void remove(int c){    L[R[c]]=L[c];    R[L[c]]=R[c];    for(int i=D[c]; i!=c; i=D[i])        for(int j=R[i]; j!=i; j=R[j])        {            U[D[j]]=U[j];            D[U[j]]=D[j];            --S[C[j]];        }}void resume(int c){    for(int i=U[c]; i!=c; i=U[i])    {        for(int j=L[i]; j!=i; j=L[j])        {            ++S[C[j]];            U[D[j]]=j;            D[U[j]]=j;        }    }    L[R[c]]=c;    R[L[c]]=c;}void init(int m)//m是列{    head=0;//头指针为0    for(int i=0; i<=m; i++)    {        U[i]=i;        D[i]=i;//建立双向十字链表        L[i]=i-1;        R[i]=i+1;        S[i]=0;    }    R[m]=0;    L[0]=m;    S[0]=INF+1;    sz=m+1;    memset(H,0,sizeof(H));}void insert(int i, int j){    if(H[i])    {        L[sz] = L[H[i]];        R[sz] = H[i];        L[R[sz]] = sz;        R[L[sz]] = sz;    }    else    {        L[sz] = sz;        R[sz] = sz;        H[i] = sz;    }    U[sz] = U[j];    D[sz] = j;    U[D[sz]] = sz;    D[U[sz]] = sz;    C[sz] = j;    ROW[sz] = i;    ++S[j];    ++sz;}bool dfs(int k,int len){    if(R[head]==head)    {        sort(O,O+len*len);        int p=0;        for(int i=0; i<len; i++)        {            for(int j=0; j<len; j++)            {                int num=O[p++];                num=num-(i*len+j)*len;                printf("%d",num);            }            puts("");        }        return  true;    }    int s=INF,c;    for (int t=R[head]; t!=head; t=R[t])        if (S[t]<s) s=S[t],c=t;    remove(c);    for(int i=D[c]; i!=c; i=D[i])    {        O[k]=ROW[i];        for(int j=R[i]; j!=i; j=R[j])            remove(C[j]);        if(dfs(k+1,len))            return  true;        for(int j=L[i]; j!=i; j=L[j])            resume(C[j]);    }    resume(c);    return  false;}void calc(int i,int j,int k,int len){    int r=(i*len+j-1)*len+k;    int base=sqrt(len);    //第i行有数字k    insert(r,i*len+k);    //第j列有数字k    insert(r,len*len+(j-1)*len+k);    //第k块有数字k    int block=(j-1)/base*base+i/base;    insert(r,len*len*2+block*len+k);    //第i行j列有一个数字(限制一个出格子只填一个数)    insert(r,len*len*3+i*len+j);}void build(char s[][10],int len)//len表示是几宫数独{    int i,j,k;    init(len*len*4);    for(i=0; i<len; i++)        for(j=1; j<=len; j++)        {            if(s[i][j-1]=='0')                for(k=1; k<=len; k++)                    calc(i,j,k,len);            else calc(i,j,s[i][j-1]-'0',len);        }}int main(){    //freopen("1.txt","r",stdin);    int t;    cin>>t;    while(t--)    {        char s[10][10];        for(int i=0; i<9; i++)            scanf("%s",s[i]);        build(s,9);        dfs(0,9);//从根开始搜    }    return 0;}

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