您的位置:首页 > 其它

hust - 1017 - Exact cover(精确覆盖DLX)

2014-10-30 11:04 363 查看
题意:N行M列的0 1矩阵(1 <= N,M <= 1000),求选出其中的一些行,使得选出的这些行每列有且仅有一个1,输出选出行的行号,无解输出"NO"。

题目链接:http://acm.hust.edu.cn/problem/show/1017

——>>DLX 练手。。

#include <cstdio>
#include <cstring>

const int MAXN = 1000 + 10;
const int MAXNODE = MAXN * MAXN;

struct DLX
{
int sz;
int H[MAXN], S[MAXN];
int row[MAXNODE], col[MAXNODE];
int U[MAXNODE], D[MAXNODE], L[MAXNODE], R[MAXNODE];
int ret[MAXN], cnt;

void Init(int n)
{
for (int i = 0; i <= n; ++i)
{
U[i] = D[i] = i;
L[i] = i - 1;
R[i] = i + 1;
}
L[0] = n;
R
= 0;

sz = n + 1;
cnt = 0;
memset(S, 0, sizeof(S));
memset(H, -1, sizeof(H));
}

void Link(const int& r, const int& c)
{
row[sz] = r;
col[sz] = c;
D[sz] = D[c];
U[D[c]] = sz;
D[c] = sz;
U[sz] = c;
if (H[r] == -1)
{
H[r] = L[sz] = R[sz] = sz;
}
else
{
R[sz] = R[H[r]];
L[R[H[r]]] = sz;
R[H[r]] = sz;
L[sz] = H[r];
}
S[c]++;
sz++;
}

void Remove(const 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[col[j]]--;
}
}
}

void Restore(const int& c)
{
for (int i = U[c]; i != c; i = U[i])
{
for (int j = L[i]; j != i; j = L[j])
{
S[col[j]]++;
U[D[j]] = j;
D[U[j]] = j;
}
}
L[R[c]] = c;
R[L[c]] = c;
}

bool Dfs(int cur)
{
if (R[0] == 0)
{
cnt = cur;
return true;
}

int c = R[0];
for (int i = R[0]; i != 0; i = R[i])
{
if (S[i] < S[c])
{
c = i;
}
}

Remove(c);
for (int i = D[c]; i != c; i = D[i])
{
ret[cur] = row[i];
for (int j = R[i]; j != i; j = R[j])
{
Remove(col[j]);
}
if (Dfs(cur + 1)) return true;
for (int j = L[i]; j != i; j = L[j])
{
Restore(col[j]);
}
}
Restore(c);

return false;
}

void Solve()
{
if (!Dfs(0))
{
puts("NO");
}
else
{
printf("%d", cnt);
for (int i = 0; i < cnt; ++i)
{
printf(" %d", ret[i]);
}
puts("");
}
}
} dlx;

int N, M;

void Read()
{
int cnt, c;

dlx.Init(M);
for (int i = 1; i <= N; ++i)
{
scanf("%d", &cnt);
while (cnt--)
{
scanf("%d", &c);
dlx.Link(i, c);
}
}
}

int main()
{
while (scanf("%d%d", &N, &M) != EOF)
{
Read();
dlx.Solve();
}

return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: