P2831 [NOIP2016 提高组] 愤怒的小鸟
抛物线从 $(0, 0)$ 开始，而三点确定一条抛物线，我们还需要两个点，也就是说，一条抛物线至少可以击中两只小猪（不考虑重复），直接枚举，将可能的抛物线记录下来。再记录这条抛物线可以击中哪些猪。$n$ 很小，考虑状态压缩，$f(x)$ 表示击中状态 $x$ 的小猪至少需要的小鸟的个数，记忆化搜索即可。注意考虑精度问题。
查看代码
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
| #include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
typedef pair <double, double> PDD;
const int N = 20;
const double eps = 1e-6;
int n, m, idx, f[1 << 20];
PDD pig[N];
struct Bird
{
double a, b;
int avai;
}birds[210];
int dfs (int x)
{
if (~f[x])
return f[x];
int res = n;
for (int i = 1; i <= idx; i++)
if ((x | birds[i].avai) != x)
res = min(res, dfs(x | birds[i].avai) + 1);
f[x] = res;
return res;
}
int main ()
{
int T;
cin >> T;
while (T--)
{
memset(birds, 0, sizeof birds);
cin >> n >> m;
idx = 0;
for (int i = 0; i < 1 << n; i++)
f[i] = -1;
f[(1 << n) - 1] = 0;
for (int i = 0; i < n; i++)
cin >> pig[i].first >> pig[i].second;
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
{
double a = (pig[i].first * pig[j].first * (pig[j].first - pig[i].first));
if (abs(a) < eps)
continue;
a = (pig[i].first * pig[j].second - pig[i].second * pig[j].first) / a;
if (a > -eps)
continue;
birds[++idx].a = a;
birds[idx].b = (pig[i].second * pig[j].first * pig[j].first - pig[j].second * pig[i].first * pig[i].first) / (pig[i].first * pig[j].first * (pig[j].first - pig[i].first));
birds[idx].avai |= (1 << i) + (1 << j);
}
for (int i = 0; i < n; i++)
for (int j = 1; j <= idx; j++)
if (abs(birds[j].a * pig[i].first * pig[i].first + birds[j].b * pig[i].first - pig[i].second) < eps)
birds[j].avai |= 1 << i;
for (int i = 0; i < n; i++)
birds[++idx].avai = (1 << i);
cout << dfs(0) << endl;
}
return 0;
}
|