bzoj 4004 装备购买

比较裸的拟阵的题，按照价值从小到大排一个序，高斯消元就行了。

这道题太变态了，我eps开1e-8就wa了，开1e-5就a了

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>

#define md
#define ll long long
#define inf (int) 1e9
#define eps 1e-5
#define N 510
using namespace std;
struct data { int id,w;} b[N];
double a[N][N];
int vis[N];
bool cmp(data a,data b) { return a.w<b.w; }
void outit()
{
	for (int i=1;i<=3;i++)
	{
		for (int j=1;j<=3;j++)
		  printf("%.2lf ",a[i][j]);
		printf("\n");
	}
	printf("\n");
}
int main()
{
	int n,m; ll ans=0; int tot=0;
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
	  for (int j=1;j<=m;j++)
	    scanf("%lf",&a[i][j]);
	for (int i=1;i<=n;i++) { scanf("%d",&b[i].w); b[i].id=i;}
	sort(b+1,b+n+1,cmp);
	for (int i=1;i<=n;i++)
	{
		int x=b[i].id;
		for (int j=1;j<=m;j++)
		  if (fabs(a[x][j])>eps)
		  {
		  	if (vis[j])
		  	{
		  		double p=a[x][j]/a[vis[j]][j];
		  		for (int k=1;k<=m;k++) a[x][k]-=p*a[vis[j]][k];
		  	}
		  	else
		  	{
		  		vis[j]=x; break;
		  	}
		 }
		for (int j=1;j<=m;j++)
		  if (fabs(a[x][j])>eps) { ans+=b[i].w; tot++; break; }
		//outit();
	}
	printf("%d %lld\n",tot,ans);
	return 0;
}