P2831 愤怒的小鸟

本来想先做天天爱跑步的, 被恶心到了, 所以来做这个题.

Solution

代码基础为XKY的爆搜代码
然后优化优化再优化.

主要做了以下优化:

• 预处理出每种状态最后一个缺失的猪.
• 预处理出打到每种猪的状态, 且状态之间不存在包含关系.
• 将打到每种猪的所有状态按打到猪的个数排序.
• 如果当前状态的答案大于等于这个状态的最优答案, 搜索无效.

然后其实最后一个优化才是最关键的, 它会减少大量的搜索.
前几种大概会让复杂度上界变小很多, 不知道该怎么描述这个东西.

#include <algorithm>
#include <iostream>
#include <cstdio>
#include <stdlib.h>
#include <cstring>
#include <cmath>
#include <vector>
using namespace std;
#pragma GCC target ("popcnt")
const double eps = 1e-7;
const int N = 55;

struct edge {
	double x, y;
} e[N];
vector<int> P[N];
int First[1 << 19];
bool vis[N][N];

inline void calc(double &a, double &b, double x1, double y1, double x2, double y2) {
	a = (x2 * y1 - x1 * y2) / (x1 * x2 * (x1 - x2));
	b = (x1 * x1 * y2 - x2 * x2 * y1) / (x1 * x2 * (x1 - x2));
}
int lim[1 << 18];
void dfs(int Status, const int& goal, int sum, int& ans) {
	if (sum >= lim[Status]) return;
	lim[Status] = min(sum, lim[Status]);
	if (Status == goal) { ans = min(ans, sum); return; }
	int p = First[Status];
	for (auto status : P[p])
		dfs(Status | status, goal, sum + 1, ans);
}

inline bool equal_double(double a, double b) {
	return a - b < eps and b - a < eps;
}
inline bool cmp(int a, int b) {
    return __builtin_popcount(a) > __builtin_popcount(b);
}

int main() {
	int T, n, k;
	scanf("%d", &T);
	while (T--) {
		scanf("%d%d", &n, &k);
		int Mx = 1 << n;
		for (int i = 0; i < Mx; i += 1) {
			for (int j = 0; j < n; j += 1)
				if (not (i & (1 << j))) {
					First[i] = j; break;
				}
		}
		memset(lim, 0x3f, sizeof lim);
		int ans = 0x7fffffff;
		memset(vis, false, sizeof vis);
		for (int i = 0; i < n; i += 1) P[i].clear();
		for (int i = 0; i < n; i += 1)
			scanf("%lf%lf", &e[i].x, &e[i].y);
		double a, b;
		int flag = false, Status = 0;
		for (int i = 0; i < n; i += 1) {
			for (int j = i + 1; j < n; j += 1) {
				if (vis[i][j]) continue;
				flag = Status = false;
				calc(a, b, e[i].x, e[i].y, e[j].x, e[j].y);
				if (a >= 0) continue;
				flag = true;
				vis[i][j] = vis[j][i] = 1;
				Status |= (1 << i) + (1 << j);
				for (int k = 0; k < n; k += 1) {
					if (k == i or k == j) continue;
					if (vis[k][i] or vis[k][j]) continue;
					if (equal_double((e[k].x * e[k].x) * a + b * e[k].x, e[k].y)) {
						Status |= (1 << k), vis[i][k] = vis[k][i] = vis[k][j] = vis[j][k] = 1;
					}
				}
				for (int k = 0; k < n; k += 1)
					if (Status & (1 << k))
					P[k].push_back(Status);
			}
			if (not flag) P[i].push_back(1 << i);
		}
		for (int i = 0; i < n; i += 1)
			std:: sort(P[i].begin(), P[i].end(), cmp);
		dfs(0, Mx - 1, 0, ans);
		printf("%d\n", ans);
	}
	return 0;
}