BZOJ 4004 [JLOI2015]装备购买 | 线性基

题目链接

Luogu P3265

题解

非常正常的线性基！

但是我不会线性基……

（吐槽：#define double long double 才过……）

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#define space putchar(' ')
#define enter putchar('\n')
typedef long long ll;
using namespace std;
template <class T>
void read(T &x){
    char c;
    bool op = 0;
    while(c = getchar(), c < '0' || c > '9')
	if(c == '-') op = 1;
    x = c - '0';
    while(c = getchar(), c >= '0' && c <= '9')
	x = x * 10 + c - '0';
    if(op) x = -x;
}
template <class T>
void write(T x){
    if(x < 0) putchar('-'), x = -x;
    if(x >= 10) write(x / 10);
    putchar('0' + x % 10);
}

#define double long double

const int N = 505;
const double eps = 1e-5;
int n, m, ins[N], ans, cnt;
struct vec {
    int val;
    double x[N];
    bool operator < (const vec &b) const{
	return val < b.val;
    }
} a[N];

void gauss(){
    for(int i = 1; i <= n; i++)
	for(int j = 1; j <= m; j++)
	    if(fabs(a[i].x[j]) > eps){
		if(!ins[j]){
		    cnt++, ans += a[i].val;
		    ins[j] = i;
		    break;
		}
		double rate = a[i].x[j] / a[ins[j]].x[j];
		for(int k = m; k >= j; k--)
		    a[i].x[k] -= rate * a[ins[j]].x[k];
	    }
}

int main(){

    read(n), read(m);
    for(int i = 1; i <= n; i++)
	for(int j = 1; j <= m; j++)
	    read(a[i].x[j]);
    for(int i = 1; i <= n; i++)
	read(a[i].val);
    sort(a + 1, a + n + 1);
    gauss();
    write(cnt), space, write(ans), enter;

    return 0;
}