Problem Description
sevenzero liked Warcraft very much, but he haven't practiced it for several years after being addicted to algorithms. Now, though he is playing with computer, he nearly losed and only his hero Pit Lord left. sevenzero is angry, he decided to cheat to turn defeat into victory. So he entered "whosyourdaddy", that let Pit Lord kill any hostile unit he damages immediately. As all Warcrafters know, Pit Lord masters a skill called Cleaving Attack and he can damage neighbour units of the unit he attacks. Pit Lord can choice a position to attack to avoid killing partial neighbour units sevenzero don't want to kill. Because sevenzero wants to win as soon as possible, he needs to know the minimum attack times to eliminate all the enemys.
Input
There are several cases. For each case, first line contains two integer N (2 ≤ N ≤ 55) and M (0 ≤ M ≤ N*N),and N is the number of hostile units. Hostile units are numbered from 1 to N. For the subsequent M lines, each line contains two integers A and B, that means A and B are neighbor. Each unit has no more than 4 neighbor units. The input is terminated by EOF.
Output
One line shows the minimum attack times for each case.
Sample Input
5 4
1 2
1 3
2 4
4 5
6 4
1 2
1 3
1 4
4 5
Sample Output
2
3
dlx重复覆盖
#include<cstdio>
#include<vector>
#include<cmath>
#include<map>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll maxn = 105;
int T, n, m, x, y, t, tot, mp[maxn][maxn];
inline void read(int &ret)
{
char c;
do {
c = getchar();
} while (c < '0' || c > '9');
ret = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
ret = ret * 10 + (c - '0');
}
struct DLX
{
#define maxn 500005
#define F(i,A,s) for (int i=A[s];i!=s;i=A[i])
int L[maxn], R[maxn], U[maxn], D[maxn];
int row[maxn], col[maxn], ans[maxn], cnt[maxn];
int n, m, num, sz;
void add(int now, int l, int r, int u, int d, int x, int y)
{
L[now] = l; R[now] = r; U[now] = u;
D[now] = d; row[now] = x; col[now] = y;
}
void reset(int n, int m)
{
num = 0x7FFFFFFF;
this->n = n; this->m = m;
for (int i = 0; i <= m; i++)
{
add(i, i - 1, i + 1, i, i, 0, i);
cnt[i] = 0;
}
L[0] = m; R[m] = 0; sz = m + 1;
}
void insert(int x, int y)
{
int ft = sz - 1;
if (row[ft] != x)
{
add(sz, sz, sz, U[y], y, x, y);
U[D[sz]] = sz; D[U[sz]] = sz;
}
else
{
add(sz, ft, R[ft], U[y], y, x, y);
R[L[sz]] = sz; L[R[sz]] = sz;
U[D[sz]] = sz; D[U[sz]] = sz;
}
++cnt[y]; ++sz;
}
//精确覆盖
void remove(int now)
{
R[L[now]] = R[now];
L[R[now]] = L[now];
F(i, D, now) F(j, R, i)
{
D[U[j]] = D[j];
U[D[j]] = U[j];
--cnt[col[j]];
}
}
void resume(int now)
{
F(i, U, now) F(j, L, i)
{
D[U[j]] = j;
U[D[j]] = j;
++cnt[col[j]];
}
R[L[now]] = now;
L[R[now]] = now;
}
bool dfs(int x)
{
//if (x + A() >= num) return;
if (!R[0]) { num = min(num, x); return true; }
int now = R[0];
F(i, R, 0) if (cnt[now]>cnt[i]) now = i;
remove(now);
F(i, D, now)
{
ans[x] = row[i];
F(j, R, i) remove(col[j]);
if (dfs(x + 1)) return true;
F(j, L, i) resume(col[j]);
}
resume(now);
return false;
}
//精确覆盖
//重复覆盖
void Remove(int now)
{
F(i, D, now)
{
L[R[i]] = L[i];
R[L[i]] = R[i];
}
}
void Resume(int now)
{
F(i, U, now) L[R[i]] = R[L[i]] = i;
}
int vis[maxn];
int flag[maxn];
int A()
{
int dis = 0;
F(i, R, 0) vis[i] = 0;
F(i, R, 0) if (!vis[i])
{
dis++; vis[i] = 1;
F(j, D, i) F(k, R, j) vis[col[k]] = 1;
}
return dis;
}
void Dfs(int x)
{
if (!R[0]) num = min(num, x);
else if (x + A()<num)
{
int now = R[0];
F(i, R, 0) if (cnt[now]>cnt[i]) now = i;
F(i, D, now)
{
Remove(i); F(j, R, i) Remove(j);
Dfs(x + 1);
F(j, L, i) Resume(j); Resume(i);
}
}
}
//重复覆盖
}dlx;
int main()
{
//read(T);
while (scanf("%d%d", &n, &m) == 2)
{
memset(mp, 0, sizeof(mp));
dlx.reset(n, n);
while (m--)
{
scanf("%d%d", &x, &y);
mp[x][y] = mp[y][x] = 1;
}
for (int i = 1; i <= n;i++)
for (int j = 1; j <= n;j++)
if (i == j || mp[i][j]) dlx.insert(i, j);
dlx.Dfs(0);
printf("%d\n", dlx.num);
}
return 0;
}
