第36届福州赛区1009 Squiggly Sudoku 解题报告
裸的DLX，比一般的数独酷稍微复杂点的就是处理输入，先dfs一下，然后建十字链表。
直接上代码了，跑得比较慢。
#include <cstdio>
#include <cstring>
using namespace std;
const int INF = 0x7fffffff;
const int NN = 350;
const int MM = 750;
int n,m; //列，行
int cntc[NN];
int L[NN*MM],R[NN*MM],U[NN*MM],D[NN*MM];
int C[NN*MM]; //列的头链表
int head;
int mx[MM][NN];
int O[MM],idx,idx2,O2[MM];
int ans[10][10];
int flags;
//删除列及其相应的行
void remove(int c)
{
int i,j;
L[R[c]] = L[c];
R[L[c]] = R[c];
for(i = D[c]; i != c; i = D[i])
{
for(j = R[i]; j != i; j = R[j])
{
U[D[j]] = U[j];
D[U[j]] = D[j];
cntc[C[j]]--;
}
}
}
//恢复列及其相应的行
void resume(int c)
{
int i,j;
R[L[c]] = c;
L[R[c]] = c;
for(i = D[c]; i != c; i = D[i])
{
for(j = R[i]; j != i; j = R[j])
{
U[D[j]] = j;
D[U[j]] = j;
cntc[C[j]]++;
}
}
}
void dfs()
{
int i,j,c;
if(flags > 1)
return;
if(R[head] == head)
{
flags++;
idx2 = idx;
for(i = 0; i < idx; i++)
O2[i] = O[i];
return;
}
int min = INF;
for(i = R[head]; i != head; i = R[i])
{
if(cntc[i] < min)
{
min = cntc[i];
c = i;
}
}
remove(c);
for(i = D[c]; i != c; i = D[i])
{
//i是某点的序号，将该点所在行的行号保存
O[idx++] = (i-1)/n;
for(j = R[i]; j != i; j = R[j])
remove(C[j]);
dfs();
if(flags > 1)
return;
for(j = L[i]; j != i; j = L[j])
resume(C[j]);
idx--;
}
resume(c);
}
void init()
{
idx = idx2 = head = 0;
flags = 0;
for (int i = 0; i <= n; i++)
{
C[i] = i;
D[i] = i;
U[i] = i;
cntc[i] = 0;
R[i] = (i+1)%(n + 1);
L[(i+1)%(n + 1)] = i;
}
}
void insert(int r, int *mk)
{
int now, pre, first;
for(int j = 0; j < 4; j++)
{
cntc[mk[j]]++;
now = r*n+mk[j];
pre = U[mk[j]];
C[now] = mk[j];
D[pre] = now;
U[now] = pre;
D[now] = mk[j];
U[mk[j]] = now;
}
pre = first = -1;
for (int i = 0; i < 4; i++)
{
now = r*n+mk[i];
if(pre == -1)
first = now;
else
{
R[pre] = now;
L[now] = pre;
}
pre = now;
}
if(first != -1)
{
R[pre] = first;
L[first] = pre;
}
}
void print()
{
int i,j;
int x,y,k;
for(i = 0; i < idx2; i++)
{
int r = O2[i];
k = r%9;
if(k==0) k = 9;
int num = (r - k)/9 + 1;
y = num%9;
if(y == 0) y = 9;
x = (num-y)/9 + 1;
ans[x][y] = k;
}
if(flags == 0)
printf("No solution\n");
else
{
if(flags > 1)
printf("Multiple Solutions\n");
else
{
for(i = 1; i <= 9; i++)
{
for(j = 1; j <= 9; j++)
printf("%d",ans[i][j]);
printf("\n");
}
}
}
}
int d[4] = {128,64,32,16};
int dir[4][2] = {{0,-1},{1,0},{0,1},{-1,0}};
int mp[10][10];
struct Point
{
bool flag[4]; //左 下 右 上
int seq;
int value;
}p[10][10];
bool change = 0;
void floodfill(int i, int j, int seq)
{
if(p[i][j].seq > 0)
return;
p[i][j].seq = seq;
for(int k = 0; k < 4; k++)
{
if(p[i][j].flag[k])
continue;
int x = i + dir[k][0];
int y = j + dir[k][1];
if(x>=1&&x<=9&&y>=1&&y<=9&&p[x][y].seq==0)
{
change = 1;
floodfill(x,y,seq);
}
}
}
int main()
{
int i,j,k;
int cases;
scanf("%d",&cases);
int tt = 0;
while(cases--)
{
for(i = 1; i <= 9; i++)
{
for(j = 1; j <= 9; j++)
{
p[i][j].seq = 0;
for(k = 0; k < 4; k++)
p[i][j].flag[k] = 0;
}
}
for(i = 1; i <= 9; i++)
for(j = 1; j <= 9; j++)
{
scanf("%d",&mp[i][j]);
for(k = 0; k < 4; k++)
{
if(mp[i][j] < 16)
{
p[i][j].value = mp[i][j];
break;
}
int t = mp[i][j]-d[k];
if(t >= 0)
{
p[i][j].flag[k] = 1;
mp[i][j] = t;
}
}
if(k == 4)
p[i][j].value = mp[i][j];
}
int seq = 1;
for(i = 1; i <= 9; i++)
{
for(j = 1; j <= 9; j++)
{
change = 0;
floodfill(i,j,seq);
if(change)
seq++;
}
}
n = 324;
m = 729;
init();
for(i = 1; i <= 9; i++)
{
for(j = 1; j <= 9; j++)
{
int t = (i-1)*9 + j;
if(p[i][j].value == 0)
{
for(k = 1; k <= 9; k++)
{
int tvec[5];
tvec[0] = t;
tvec[1] = (81+(i-1)*9+k);
tvec[2] = (162+(j-1)*9+k);
tvec[3] = (243+(p[i][j].seq-1)*9+k);
insert(9*(t-1)+k, tvec);
}
}
else
{
int tvec[5];
k = p[i][j].value;
tvec[0] = t;
tvec[1] = (81+(i-1)*9+k);
tvec[2] = (162+(j-1)*9+k);
tvec[3] = (243+(p[i][j].seq-1)*9+k);
insert(9*(t-1)+k, tvec);
}
}
}
printf("Case %d:\n",++tt);
dfs();
print();
}
return 0;
}