题意：N(2 ≤ N ≤ 55)个点，M(0 ≤ M ≤ N*N)条无向边，删除一个点会把与其相邻的点一起删掉。问最少删几次能够删掉全部点。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3498

——>>N个点看成 N 个要被覆盖的列，每一个点作为一行，与其相邻的点的位置在这一行中标为 1，还有它自已的位置也标记为 1。。

这就是经典的反复覆盖问题了。。于是，DLX上场。。

#include <cstdio>
#include <cstring>const int MAXR = 55 + 10;
const int MAXC = 55 + 10;
const int MAXNODE = MAXR * MAXC;
const int INF = 0x3f3f3f3f;struct DLX
{int sz;int H[MAXR], S[MAXC];int row[MAXNODE], col[MAXNODE];int U[MAXNODE], D[MAXNODE], L[MAXNODE], R[MAXNODE];int Min;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[n] = 0;sz = n + 1;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){for (int i = D[c]; i != c; i = D[i]){L[R[i]] = L[i];R[L[i]] = R[i];}}void Restore(const int& c){for (int i = U[c]; i != c; i = U[i]){L[R[i]] = i;R[L[i]] = i;}}int A(){int ret = 0;bool vis[MAXC];memset(vis, 0, sizeof(vis));for (int i = R[0]; i != 0; i = R[i]){if (!vis[i]){vis[i] = true;++ret;for (int j = D[i]; j != i; j = D[j]){for (int k = R[j]; k != j; k = R[k]){vis[col[k]] = true;}}}}return ret;}void Dfs(int cur){if (cur + A() >= Min) return;if (R[0] == 0){if (cur < Min){Min = cur;}return;}int c = R[0];for (int i = R[0]; i != 0; i = R[i]){if (S[i] < S[c]){c = i;}}for (int i = D[c]; i != c; i = D[i]){Remove(i);for (int j = R[i]; j != i; j = R[j]){Remove(j);}Dfs(cur + 1);for (int j = L[i]; j != i; j = L[j]){Restore(j);}Restore(i);}}void Solve(){Min = INF;Dfs(0);printf("%d\n", Min);}} dlx;int N, M;void Read()
{int a, b;dlx.Init(N);while (M--){scanf("%d%d", &a, &b);dlx.Link(a, b);dlx.Link(b, a);}for (int i = 1; i <= N; ++i){dlx.Link(i, i);}
}int main()
{while (scanf("%d%d", &N, &M) == 2){Read();dlx.Solve();}return 0;
}