POJ 2676 3074 3076 数独3题 Dancing Links
今天就一直在做DLX的题目了，目前也只是入门而已，就做了几个数独的问题
将数独转化为DLX的模型是很繁琐的一部分。
首先，必须明白，有多少种选择就要建立多少行，有多少个条件约束就要建立多少个列
对于一个9*9的数独，我们可以在81个位置中选择一个地方，放1~9的数，所以选择数目就是9*9*9了，就要建立这么些个行
而约束呢，首先是81个格子，每个格子有且只能有1个数，那么就是9*9个列，然后对于某个位置的某个数，同一行的不能再有一样的数，也是9*9个列，同一列，同一宫都是一个原理，所以总共就是4*9*9个列了
然后问题就转化为了，从9*9*9行中选择某些行，使得每一列都只有1个1，这里的1代表的就是唯一性，比如每一宫中只能有1个1，如果你选的两行致使1个宫内有两个1，那么显然这是不符合约束的，必然在某一列中这个冲突会得到体现。同样的是同一列和同一行。
对于每个格子有且只能有1个数这个条件，也是不可少的。虽说插入的时候只要放置的位置不同，一定不会产生冲突，但是加入此条件就能保证最后输出的结果，每一格都充满了数字。
然后建图的时候就是按着行和列用十字循环链表建图了。
怎么说呢，今天这三道题做的，感觉学了不少东西，但是写起来，却又写挫了，16*16那个数独也不知他们怎么写的，我花了300+ms，前边一堆100ms以内的
首先是2676
/*
ID: sdj22251
PROG: inflate
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define MAXN 9*9*9*9
#define INF 1000000000
#define N 9
#define M 9*9*9+5
using namespace std;
int L[MAXN], R[MAXN], C[MAXN], S[M], U[MAXN], D[MAXN], H[M], O[M], X[MAXN]; //X用来存储行，C用来存储列，H用来存储每行中的第一个结点，O用来存储结果
int cnt, head;
char mp[N + 5][N + 5], ans[N * N + 5];
bool vis[N * N * 4 + 5];
void link(int r, int c)
{
S[c]++;
C[cnt] = c;
X[cnt] = r;
U[cnt] = c;
D[cnt] = D[c];
U[D[c]] = cnt;
D[c] = cnt;
if(H[r] == -1)
{
H[r] = cnt;
L[cnt] = R[cnt] = cnt;
}
else
{
L[cnt] = H[r];
R[cnt] = R[H[r]];
L[R[H[r]]] = cnt;
R[H[r]] = cnt;
}
cnt++;
}
void init()
{
cnt = 0;
head = 0;
for(int i = 0; i <= N * N * 4; i++)
{
S[i] = 0;
vis[i] = 0;
D[i] = U[i] = i;
R[i] = (i + 1) % (N * N * 4 + 1);
L[i] = (i + N * N * 4) % (N * N * 4 + 1);
cnt++;
}
memset(H, -1, sizeof(H));
}
void cal(int &r, int &cx, int &cy, int &ck, int &cg, int i, int j, int k)
{
r = (i * N + j) * N + k - 1; //代表所属的行
cg = i * N + j + 1; //代表的是数独中i,j位置所属的列
cx = N * N + i * N + k; //代表数独中同一行所属的列
cy = N * N * 2 + j * N + k; //代表数独中同一列所属的列
int LN = (int)sqrt(N * 1.0);
ck = N * N * 3 + (i / LN * LN + j / LN) * N + k;//代表数独中同一宫所属的列
}
void readdata()
{
int r, cx, cy, ck, cg;
for(int i = 0; i < N; i++)
scanf("%s", mp[i]);
for(int i = 0; i < N; i++)
for(int j = 0; j < N; j++)
if(mp[i][j] != '0')
{
cal(r, cx, cy, ck, cg, i, j, mp[i][j] - '0');
link(r, cx);
link(r, cy);
link(r, ck);
link(r, cg);
vis[cx] = vis[cy] = vis[ck] = vis[cg] = 1;
}
for(int i = 0; i < N; i++)
for(int j = 0; j < N; j++)
for(int k = 1; k <= N; k++)
{
cal(r, cx, cy, ck, cg, i, j, k);
if(vis[cx] || vis[cy] || vis[ck] || vis[cg]) continue;
link(r, cx);
link(r, cy);
link(r, ck);
link(r, cg);
}
}
void removes(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 resumes(int c)
{
for(int i = U[c]; i != c; i = U[i])
for(int j = L[i]; j != i; j = L[j])
{
U[D[j]] = j;
D[U[j]] = j;
S[C[j]]++;
}
L[R[c]] = c;
R[L[c]] = c;
}
bool dfs(int k)
{
if(R[head] == head)
{
for(int i = 0; i < k; i++) //根据所选择的行 能推出选择的位置和数字
ans[X[O[i]] / N] = X[O[i]] % 9 + '1';
for(int i = 0; i < N * N; i++)
{
printf("%c", ans[i]);
if((i + 1) % N == 0) printf("\n");
}
return true;
}
int s = INF, c;
for(int i = R[head]; i != head; i = R[i])
if(s > S[i])
{
s = S[i];
c = i;
}
removes(c);
for(int i = U[c]; i != c; i = U[i])
{
O[k] = i;
for(int j = R[i]; j != i; j = R[j])
removes(C[j]);
if(dfs(k + 1)) return true;
for(int j = L[i]; j != i; j = L[j])
resumes(C[j]);
}
resumes(c);
return false;
}
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
init();
readdata();
dfs(0);
}
return 0;
}
然后是3074
/*
ID: sdj22251
PROG: inflate
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define MAXN 9*9*9*9
#define INF 1000000000
#define N 9
#define M 9*9*9+5
using namespace std;
int L[MAXN], R[MAXN], C[MAXN], S[M], U[MAXN], D[MAXN], H[M], O[M], X[MAXN];
int cnt, head;
char mp[N * N + 5], ans[N * N + 5];
bool vis[N * N * 4 + 5];
void link(int r, int c)
{
S[c]++;
C[cnt] = c;
X[cnt] = r;
U[cnt] = c;
D[cnt] = D[c];
U[D[c]] = cnt;
D[c] = cnt;
if(H[r] == -1)
{
H[r] = cnt;
L[cnt] = R[cnt] = cnt;
}
else
{
L[cnt] = H[r];
R[cnt] = R[H[r]];
L[R[H[r]]] = cnt;
R[H[r]] = cnt;
}
cnt++;
}
void init()
{
cnt = 0;
head = 0;
for(int i = 0; i <= N * N * 4; i++)
{
S[i] = 0;
vis[i] = 0;
D[i] = U[i] = i;
R[i] = (i + 1) % (N * N * 4 + 1);
L[i] = (i + N * N * 4) % (N * N * 4 + 1);
cnt++;
}
memset(H, -1, sizeof(H));
}
void cal(int &r, int &cx, int &cy, int &ck, int &cg, int i, int j, int k)
{
r = (i * N + j) * N + k - 1;
cg = i * N + j + 1;
cx = N * N + i * N + k;
cy = N * N * 2 + j * N + k;
int LN = (int)sqrt(N * 1.0);
ck = N * N * 3 + (i / LN * LN + j / LN) * N + k;
}
void readdata()
{
int r, cx, cy, ck, cg;
for(int i = 0; i < N; i++)
for(int j = 0; j < N; j++)
if(mp[i * N + j] != '.')
{
cal(r, cx, cy, ck, cg, i, j, mp[i * N + j] - '0');
link(r, cx);
link(r, cy);
link(r, ck);
link(r, cg);
vis[cx] = vis[cy] = vis[ck] = vis[cg] = 1;
}
for(int i = 0; i < N; i++)
for(int j = 0; j < N; j++)
if(mp[i * N + j] == '.')
for(int k = 1; k <= N; k++)
{
cal(r, cx, cy, ck, cg, i, j, k);
if(vis[cx] || vis[cy] || vis[ck] || vis[cg]) continue;
link(r, cx);
link(r, cy);
link(r, ck);
link(r, cg);
}
}
void removes(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 resumes(int c)
{
for(int i = U[c]; i != c; i = U[i])
for(int j = L[i]; j != i; j = L[j])
{
U[D[j]] = j;
D[U[j]] = j;
S[C[j]]++;
}
L[R[c]] = c;
R[L[c]] = c;
}
bool dfs(int k)
{
if(R[head] == head)
{
for(int i = 0; i < k; i++)
ans[X[O[i]] / N] = X[O[i]] % N + '1';
for(int i = 0; i < N * N; i++)
{
printf("%c", ans[i]);
}
printf("\n");
return true;
}
int s = INF, c;
for(int i = R[head]; i != head; i = R[i])
if(s > S[i])
{
s = S[i];
c = i;
}
removes(c);
for(int i = U[c]; i != c; i = U[i])
{
O[k] = i;
for(int j = R[i]; j != i; j = R[j])
removes(C[j]);
if(dfs(k + 1)) return true;
for(int j = L[i]; j != i; j = L[j])
resumes(C[j]);
}
resumes(c);
return false;
}
int main()
{
while(scanf("%s", mp) != EOF && mp[0] != 'e')
{
init();
readdata();
dfs(0);
}
return 0;
}
最后是3076
这是16*16的数独
/*
ID: sdj22251
PROG: inflate
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define MAXN 16*16*16*16+16*16*4
#define INF 1000000000
#define N 16
#define M 16*16*16+5
using namespace std;
int L[MAXN], R[MAXN], C[MAXN], S[M], U[MAXN], D[MAXN], H[M], O[M], X[MAXN];
int cnt, head;
char mp[N + 5][N + 5], ans[N * N + 5];
bool vis[N * N * 4 + 5];
void link(int r, int c)
{
S[c]++;
C[cnt] = c;
X[cnt] = r;
U[cnt] = c;
D[cnt] = D[c];
U[D[c]] = cnt;
D[c] = cnt;
if(H[r] == -1)
{
H[r] = cnt;
L[cnt] = R[cnt] = cnt;
}
else
{
L[cnt] = H[r];
R[cnt] = R[H[r]];
L[R[H[r]]] = cnt;
R[H[r]] = cnt;
}
cnt++;
}
void init()
{
cnt = 0;
head = 0;
for(int i = 0; i <= N * N * 4; i++)
{
S[i] = 0;
vis[i] = 0;
D[i] = U[i] = i;
R[i] = (i + 1) % (N * N * 4 + 1);
L[i] = (i + N * N * 4) % (N * N * 4 + 1);
cnt++;
}
memset(H, -1, sizeof(H));
}
void cal(int &r, int &cx, int &cy, int &ck, int &cg, int i, int j, int k)
{
r = (i * N + j) * N + k - 1;
cg = i * N + j + 1;
cx = N * N + i * N + k;
cy = N * N * 2 + j * N + k;
int LN = (int)sqrt(N * 1.0);
ck = N * N * 3 + (i / LN * LN + j / LN) * N + k;
}
void readdata()
{
int r, cx, cy, ck, cg;
for(int i = 1; i < N; i++)
scanf("%s", mp[i]);
for(int i = 0; i < N; i++)
for(int j = 0; j < N; j++)
if(mp[i][j] != '-')
{
cal(r, cx, cy, ck, cg, i, j, mp[i][j] - 'A' + 1);
link(r, cx);
link(r, cy);
link(r, ck);
link(r, cg);
vis[cx] = vis[cy] = vis[ck] = vis[cg] = 1;
}
for(int i = 0; i < N; i++)
for(int j = 0; j < N; j++)
if(mp[i][j] == '-')
for(int k = 1; k <= N; k++)
{
cal(r, cx, cy, ck, cg, i, j, k);
if(vis[cx] || vis[cy] || vis[ck] || vis[cg]) continue;
link(r, cx);
link(r, cy);
link(r, ck);
link(r, cg);
}
}
void removes(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 resumes(int c)
{
for(int i = U[c]; i != c; i = U[i])
for(int j = L[i]; j != i; j = L[j])
{
U[D[j]] = j;
D[U[j]] = j;
S[C[j]]++;
}
L[R[c]] = c;
R[L[c]] = c;
}
bool dfs(int k)
{
if(R[head] == head)
{
for(int i = 0; i < k; i++)
ans[X[O[i]] / N] = X[O[i]] % N + 'A';
for(int i = 0; i < N * N; i++)
{
printf("%c", ans[i]);
if((i + 1) % N == 0) printf("\n");
}
printf("\n");
return true;
}
int s = INF, c;
for(int i = R[head]; i != head; i = R[i])
if(s > S[i])
{
s = S[i];
c = i;
}
removes(c);
for(int i = U[c]; i != c; i = U[i])
{
O[k] = i;
for(int j = R[i]; j != i; j = R[j])
removes(C[j]);
if(dfs(k + 1)) return true;
for(int j = L[i]; j != i; j = L[j])
resumes(C[j]);
}
resumes(c);
return false;
}
int main()
{
while(scanf("%s", mp[0]) != EOF)
{
init();
readdata();
dfs(0);
}
return 0;
}