POJ 3076 Sudoku 精确覆盖问题DLX

题目：

题意：

思路

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

const int X = 300000 + 10, N = 5000 + 10, M = 1200 + 10, INF = 0x3f3f3f3f;

struct DLX
{
int U[X], D[X], L[X], R[X], row[X], col[X];
int H
, S[M];
int head, sz, tot, n, m, ans
;
void init(int _n, int _m)
{
n = _n, m = _m;
for(int i = 0; i <= m; i++)
L[i] = i-1, R[i] = i+1, U[i] = D[i] = i, S[i] = 0;
head = 0, tot = 0, sz = m;
L[head] = m, R[m] = head;
for(int i = 1; i <= n; i++) H[i] = -1;
}
void link(int r, int c)
{
++S[col[++sz]=c];
row[sz] = r;
D[sz] = D[c], U[D[c]] = sz;
U[sz] = c, D[c] = sz;
if(H[r] < 0) H[r] = L[sz] = R[sz] = sz;
else R[sz] = R[H[r]], L[R[H[r]]] = sz, L[sz] = H[r], R[H[r]] = sz;
}
void del(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])
D[U[j]] = D[j], U[D[j]] = U[j], --S[col[j]];
}
void recover(int c)
{
for(int i = U[c]; i != c; i = U[i])
for(int j = L[i]; j != i; j = L[j])
D[U[j]] = U[D[j]] = j, ++S[col[j]];
R[L[c]] = L[R[c]] = c;
}
bool dance(int dep)
{
if(R[head] == head)
{
tot = dep-1; return true;
}
int c = R[head];
for(int i = R[head]; i != head; i = R[i])
if(S[i] < S[c]) c = i;
del(c);
for(int i = D[c]; i != c; i = D[i])
{
ans[dep] = row[i];
for(int j = R[i]; j != i; j = R[j]) del(col[j]);
if(dance(dep + 1)) return true;
for(int j = L[i]; j != i; j = L[j]) recover(col[j]);
}
recover(c);
return false;
}
}dlx;
int encode(int a, int b, int c, int len)
{
return a*len*len + b*len + c;
}
void calc(int x, int y, int k, int len)
{
int r = ((x-1) * len + (y-1)) * len + k;
dlx.link(r, encode(0, x-1, y, len));
dlx.link(r, encode(1, x-1, k, len));
dlx.link(r, encode(2, y-1, k, len));

int base = 4;
int block = (x-1) / base * base + (y-1) / base + 1;
dlx.link(r, encode(3, block-1, k, len));
//    dlx.link(r, (x-1) * len + y);
//    dlx.link(r, len*len + (x-1) * len + k);
//    dlx.link(r, len*len*2 + (y-1) * len + k);
//
//    int base = 4;
//    int block = (x-1) / base * base + (y-1) / base + 1;
//    dlx.link(r, len*len*3 + (block-1) * len + k);
}
//有两个不同的输出函数，用哪一个都可以，之前用的注释掉的这个，这次用另外一个
//void print(int len)
//{
//    sort(dlx.ans + 1, dlx.ans + 1 + dlx.tot);
//    for(int i = 1; i <= len; i++)
//    {
//        for(int j = 1; j <= len; j++)
//            printf("%c", dlx.ans[(i-1)*len+j] - ((i-1)*len + (j-1)) * len - 1 + 'A');
//        printf("\n");
//    }
//}
void recode(int val, int &a, int &b, int &c)
{
val--;
c = val % 16; val /= 16;
b = val % 16; val /= 16;
a = val;
}
void print(int len, char str[][20])
{
int x, y, k;
for(int i = 1; i <= dlx.tot; i++)
{
recode(dlx.ans[i], x, y, k);
str[x+1][y+1] = 'A' + k;
}
for(int i = 1; i <= len; i++) printf("%s\n", str[i] + 1);
}
int main()
{
int len = 16, cas = 0;
char str[20][20];
while(~ scanf("%s", str[1]+1))
{
for(int i = 2; i <= len; i++) scanf("%s", str[i]+1);
if(cas++ != 0) printf("\n");
dlx.init(len * len * len, len * len * 4);
for(int i = 1; i <= len; i++)
for(int j = 1; j <= len; j++)
if(str[i][j] == '-')
for(int k = 1; k <= len; k++) calc(i, j, k, len);
else calc(i, j, str[i][j]-'A'+1, len);
dlx.dance(1);
//print(len);
print(len, str);
}
return 0;
}