P3265 [JLOI2015]装备购买
线性基可以用高斯消元做，为了简化这一过程，所以有了这样的“基”，如果不存在这样的基，添加即可，否则与当前的基消元。
查看代码
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| #include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
#define double long double
template <class Type>
void read(Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
flag && (x = ~x + 1);
}
template <class Type>
void write(Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar(x % 10 + '0');
}
const double eps = 1e-8;
const int N = 510;
struct Node
{
int c;
vector <double> v;
bool operator < (const Node &_) const
{
return c < _.c;
}
} w[N];
int n, m, p[N];
int sign (double x)
{
if (x > eps)
return 1;
if (x < -eps)
return -1;
return 0;
}
int main ()
{
read(n), read(m);
for (int i = 0; i < n; i++)
for (int j = 0, a; j < m; j++)
{
read(a);
w[i].v.push_back(a);
}
for (int i = 0; i < n; i++)
read(w[i].c);
sort(w, w + n);
for (int i = 0; i < m; i++)
p[i] = -1;
int cnt = 0, cst = 0;
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
{
if (!sign(w[i].v[j]))
continue;
if (p[j] == -1)
{
p[j] = i;
cnt++, cst += w[i].c;
break;
}
else
{
double t = w[i].v[j] / w[p[j]].v[j];
for (int k = j; k < m; k++)
w[i].v[k] -= t * w[p[j]].v[k];
}
}
printf("%d %d", cnt, cst);
return 0;
}
|