洛谷 [P3265] 装备购买

线性基

通过题目描述可以感觉到就是要求线性基,

线性基的求法是高斯消元,消完以后剩下的x的系数非 0 的就是线性基

本题有一个贪心策略,每次挑选价格最小的来消掉其他的元

//可以快排预处理
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const int MAXN = 505;
int b[MAXN], n, m;
long double a[MAXN][MAXN];
int main() {
	cin >> n >> m;
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= m; j++) {
			cin >> a[i][j];
		}
	}
	for(int i = 1; i <= n; i++) cin >> b[i];
	for(int i = 1; i <= m; i++) {
		for(int j = i + 1; j <= n; j++) {
			if(fabs(a[j][i]) > 1e-4 && b[j] < b[i]) {
				for(int k = 1; k <= n; k++) swap(a[i][k], a[j][k]);
				swap(b[i], b[j]);
			}
		}
		if(!a[i][i]) continue;
		for(int j = 1; j <= n; j++) {
			if(i == j) continue;
			long double rate = (a[j][i] / a[i][i]);
			for(int k = i; k <= m; k++) {
				a[j][k] -= a[i][k] * rate;
			}
		}
	}
	int cnt = 0, ans = 0;
	for(int i = 1; i <= min(n, m); i++) {
		if(fabs(a[i][i]) > 1e-4) cnt++, ans += b[i];
	}
	printf("%d %d\n", cnt, ans);
	return 0;
}

线性基优化版

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#define eps 1e-5
using namespace std;
const int MAXN = 505;
int n, m, nxt[MAXN], cnt, ans;
struct qwq{
	long double num[MAXN];
	int val;
	bool operator < (const qwq &b) const{
		return this -> val < b.val;
	}
}a[MAXN];
bool f[MAXN];
int main() {
	cin >> n >> m;
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= m; j++) {
			cin >> a[i].num[j];
		}
	}
	for(int i = 1; i <= n; i++) cin >> a[i].val;
	sort(a + 1, a + n + 1);
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= m; j++) {
			if(fabs(a[i].num[j]) > eps){
				if(!f[j]) {
					f[j] = 1;
					cnt++;
					ans += a[i].val;
					nxt[j] = i;
					break;
				}else{
					long double rate = a[i].num[j] / a[nxt[j]].num[j];
					for(int k = j; k <= n; k++) a[i].num[k] -= rate * a[nxt[j]].num[k];
				}
			}
		}
	}
	printf("%d %d\n", cnt, ans);
	return 0;
}