POJ 3740 Easy Finding（舞蹈链）

/*
舞蹈链模板题
*/

#include <cstdio>
#include <cstring>
const int mMax = 50, nMax = 500;
int map[mMax][nMax];
int M, N;
struct Node
{
int left, right, up, down;
int col;
}node[mMax * nMax];//双向链表
int len;
int cnt[nMax];

void init(int n)
{
int i;
for(i = 0; i <= n; ++ i)
{
node[i].left = i - 1;
node[i].right = i + 1;
node[i].col = i;//
node[i].down = i;
node[i].up = i;
}
node
.right = 0;
node[0].left = n;
}

void remove(int c)
{
int i, j;
node[node[c].left].right = node[c].right;
node[node[c].right].left = node[c].left;
for(i = node[c].down; i != c; i = node[i].down)
{
for(j = node[i].right; j != i; j = node[j].right)
//因为每列只能包含一个1，所以可以把所有的列中对应的行全部删掉。
//这里还有一个细节，因为前面执行过删除列操作，所以不需要对i再执行删除行操作，因为以后肯定不会再访问到
{
node[node[j].up].down = node[j].down;
node[node[j].down].up = node[j].up;
cnt[node[j].col] --;
}
}
}

void resume(int c)//先删除的后恢复
{
int i, j;
for(i = node[c].up; i != c; i = node[i].up)
{
for(j = node[i].left; j != i; j = node[j].left)
{
node[node[j].up].down = j;
node[node[j].down].up = j;
cnt[node[j].col] ++;
}
}
node[node[c].left].right = c;
node[node[c].right].left = c;

//先删除先恢复，也正确，但是有些情况会出现偏差，无法与原来状态对称
/*int i, j;
node[node[c].left].right = c;
node[node[c].right].left = c;
for(i = node[c].down; i != c; i = node[i].down)
{
for(j = node[i].right; j != i; j = node[j].right)
{
node[node[j].up].down = j;
node[node[j].down].up = j;
cnt[node[j].col] ++;
}
}*/
}

bool dfs(int k)//k表示已经找出了多少行
{
int i, j;
if(node
.right == N)//终止状态：列已经被全部删除。这里需要一个空节点，来进行判断
return true;
int c, _min = 0x7fffffff;
for(i = node
.right; i != N; i = node[i].right)
{
if(cnt[i] < _min)
{
_min = cnt[i];
c = i;
}
}
remove(c);
for(i = node[c].down; i != c; i = node[i].down)
{
//找到"一行"可能的解，然后递归
for(j = node[i].right; j != i; j = node[j].right)
{
remove(node[j].col);
}
if(dfs(k + 1)) return true;
for(j = node[i].right; j != i; j = node[j].right)
{
resume(node[j].col);
}
}
resume(c);
return false;
}

int main()
{
//freopen("e://data.in", "r", stdin);
while(scanf("%d%d", &M, &N) != EOF)
{
int i, j;
init(N);
int cur;
memset(cnt, 0, sizeof(cnt));
cur = N + 1;
for(i = 0; i < M; ++ i)
{
int start = cur;
node[start].left = start;//初始化，为了形成环
node[start].right = start;
for(j = 0; j < N; ++ j)
{
scanf("%d", &map[i][j]);
if(map[i][j] == 1)
{
node[cur].right = node[start].right;//这里需要四步操作，原来漏泄了③，一直WA。模拟双向链表的头部插入操作
node[cur].left = start;
node[node[start].right].left = cur;//③
node[start].right = cur;

node[cur].down = node[j].down;
node[cur].up = j;
node[node[j].down].up = cur;
node[j].down = cur;

node[cur].col = j;
cur ++;
cnt[j] ++;
}
}
}
if(dfs(0))
printf("Yes, I found it\n");
else
printf("It is impossible\n");
}
return 0;
}