poj 3740 Easy Finding

3 3
0 1 0
0 0 1
1 0 0
4 4
0 0 0 1
1 0 0 0
1 1 0 1
0 1 0 0

Yes, I found it
It is impossible

#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
#include <algorithm>
#define maxn 5200
#define maxl 305
#define maxm 18
#define head 0

//数组实现的dancing links
int up[maxn] = {0}, down[maxn] = {0}, left[maxn] = {0}, right[maxn] = {0};//顾名思义是某个结点的上下左右结点的标号
int column[maxn] = {0}, row[maxn] = {0}, ans[maxn] = {0};//某个结点对应的行号,ans是取的行号的数组
int m = 0, n = 0;

void init()//初始化
{
memset(up, 0, sizeof(up));
memset(down, 0, sizeof(down));
memset(left, 0, sizeof(left));
memset(right, 0, sizeof(right));
memset(column, 0, sizeof(column));
memset(row, 0, sizeof(row));
memset(ans, 0, sizeof(ans));

return;
}

//这里的remove其实并没有真正删除掉结点,可以用resume恢复
void remove(int c)//去掉c号结点,以及在c所在列上有结点的行的结点
{
left[right[c]] = left[c];
right[left[c]] = right[c];

for (int i = down[c]; i != c; i = down[i])
{
for (int j = right[i]; j != i; j = right[j])
{
up[down[j]] = up[j];
down[up[j]] = down[j];
}
}
return;
}

void resume(int c)//remove的逆操作
{
left[right[c]] = c;
right[left[c]] = c;
for (int i = up[c]; i != c; i = up[i])
{
for (int j = right[i]; j != i; j = right[j])
{
up[down[j]] = j;
down[up[j]] = j;
}
}
return;
}

//我们的终极目标实际上是把十字链表中所有的结点都删掉,只留一个头结点
bool dance(int k)//这里的k实际上是没多大意义的
{
int c = right[head];
if (c == head)//只剩下十字链表的头结点,则完成了目标
{
return true;
}

remove(c);
for (int i = down[c]; i != c; i = down[i])//这里其实是枚举某一行对待匹配行的第c位进行匹配
{
ans[k] = row[i];
for (int j = right[i]; j != i; j = right[j])
remove(column[j]);//remove待匹配行的column[j]结点
if (dance(k + 1))
return true;
for (int j = left[i]; j != i; j = left[j])
resume(column[j]);
}
resume(c);
return false;
}

void test(int idn)//测试用
{
printf("up do le ri co ro\n");
for (int i = 0; i < idn; ++i)
{
printf("%02d %02d %02d %02d %02d %02d\n", up[i], down[i], left[i], right[i], column[i], row[i]);
}
return;
}

int main()
{

while (scanf("%d%d", &m, &n) != EOF)//输入部分极其复杂
{
init();
int id = 1;
int tmp = 0;
int lineup[maxl] = {0};

for (int j = 0; j < n; ++j)//待匹配行
{
lineup[j] = id;
column[id] = j + 1;
if (j != n - 1)
{
right[id] = id + 1;
}
left[id] = id - 1;
++id;
}
right[head] = 1;
left[head] = id - 1;
right[id - 1] = head;

for (int i = 0; i < m; ++i)
{
int first_id = 0;
int left_id = 0;
for (int j = 0; j < n; ++j)
{
scanf("%d", &tmp);
if (tmp)
{
row[id] = i + 1;
column[id] = j + 1;
if (first_id == 0)
{
first_id = id;
left_id = id;
}
else
{
left[id] = left_id;
right[left_id] = id;
left_id = id;
}
int upid = lineup[j];
up[id] = upid;
down[upid] = id;
lineup[j] = id;
++id;
}
}
if (first_id != 0)
{
left[first_id] = left_id;
right[left_id] = first_id;
}
}

for (int j = 0; j < n; ++j)
{
down[lineup[j]] = j + 1;
up[j + 1] = lineup[j];
}

//test(id);

int is_possible = dance(0);

if (is_possible)
printf("Yes, I found it\n");
else
printf("It is impossible\n");

}
return 0;
}