标签:log printf pre nbsp ble include 卡精度 lan tin
感觉这题不是很难,但是很恶心。
说一下几点。
1.预处理出来每两个点所构成的抛物线能消除的猪的集合。
2.如果两个点横坐标相同,则不能构成抛物线
3.a >= 0 continue
4.卡精度
5.卡常数(本蒟蒻巨菜,2nn2做法)
#include <cstdio>
#include <cstring>
#define N 19
#define abs(x) ((x) < 0 ? -(x) : (x))
#define min(x, y) ((x) < (y) ? (x) : (y))
int T, n, m, S;
int f[1 << N], s[N][N];
double X[N], Y[N], a, b;
inline bool pd(double x, double y)
{
return abs(x - y) < (1e-6);
}
int main()
{
int i, j, k, l;
scanf("%d", &T);
while(T--)
{
scanf("%d %d", &n, &m);
memset(f, 127 / 3, sizeof(f));
for(i = 1; i <= n; i++) scanf("%lf %lf", &X[i], &Y[i]);
memset(s, 0, sizeof(s));
for(i = 1; i <= n; i++)
for(j = i + 1; j <= n; j++)
{
if(pd(X[i], X[j])) continue;
a = (Y[j] / X[j] - Y[i] / X[i]) / (X[j] - X[i]);
b = Y[i] / X[i] - a * X[i];
if(a >= 0) continue;
s[i][j] |= (1 << i - 1) | (1 << j - 1);
for(k = 1; k <= n; k++)
if(k != i && k != j && pd(Y[k], a * X[k] * X[k] + b * X[k]))
s[i][j] |= 1 << k - 1;
}
f[0] = 0;
for(i = 0; i < (1 << n); i++)
for(j = 1; j <= n; j++)
if(!(i & (1 << j - 1)))
{
f[i | (1 << j - 1)] = min(f[i | (1 << j - 1)], f[i] + 1);
for(k = j + 1; k <= n; k++)
if((i & (1 << k - 1)) && s[j][k])
{
S = i ^ (i & s[j][k]);
f[i | (1 << j - 1)] = min(f[i | (1 << j - 1)], f[S] + 1);
}
}
printf("%d\n", f[(1 << n) - 1]);
}
return 0;
}
标签:log printf pre nbsp ble include 卡精度 lan tin
原文地址:http://www.cnblogs.com/zhenghaotian/p/7354387.html