UVA 11008 Antimatter Ray Clearcutting

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明显的状态压缩DP，状态转移时要分打掉一个点或者打掉一条线，用最朴素的转移，总时间复杂度是O(2^N*N^3)，姿势优越的话可以水过这题。

//11008 Antimatter Ray Clearcutting Accepted C++ 2.656 2012-12-12 02:15:41#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int MAX = 16;const int INF = 0x3f3f3f3f;int T, N, M, X[MAX], Y[MAX];int cnt[1<<MAX], d[1<<MAX];void dfs(int x, int y, int z){if (x == MAX){cnt[z] = y;return;}z &= (~(1<<x));dfs(x+1, y, z);z |= (1<<x);dfs(x+1, y+1, z);}int main(){dfs(0, 0, 0);scanf("%d", &T);for (int cas = 1; cas <= T; cas++){scanf("%d%d", &N, &M);int tot = 1<<N;for (int i = 0; i < N; i++)scanf("%d%d", &X[i], &Y[i]);fill(d+1, d+tot, INF);int ans = INF;for (int u = 0; u < tot; u++) if (d[u] < INF){if (cnt[u] >= M)ans = min(ans, d[u]);for (int i = 0; i < N; i++) if (!((1<<i)&u)){int v = u;for (int j = 0; j < N; j++)if (X[i] == X[j] && Y[i] == Y[j])v |= (1<<j);d[v] = min(d[v], d[u]+1);for (int j = i+1; j < N; j++) if (!((1<<j)&u))if (!(X[i] == X[j] && Y[i] == Y[j])){int v = u;for (int k = 0; k < N; k++)if ((X[i]-X[j])*(Y[j]-Y[k]) == (X[j]-X[k])*(Y[i]-Y[j]))v |= (1<<k);d[v] = min(d[v], d[u]+1);}}}if (cas > 1)printf("\n");printf("Case #%d:\n%d\n", cas, ans);}return 0;}