hdu - 3498 - whosyourdaddy(重复覆盖DLX)
2014-10-30 18:04
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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,问最少删几次可以删掉所有点。
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。。
这就是经典的重复覆盖问题了。。于是,DLX上场。。
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。。
这就是经典的重复覆盖问题了。。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 = 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; }
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