高斯消元

装备购买HYSBZ - 4004 

程序使用long double 这一类型，相比于double,long double 的精度更高，但运算速度稍微慢点

这是一道及其典型的高斯消元，应对的也是各种情况，有解的，没解的，已经未知数个数比方程数多或者少或者相等

可以当做模板

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<queue>
#include<map>
#include<set>
#include<list>
#include<ctime>
#include<ctype.h>
#include<bitset>
#include<algorithm>
#include<numeric> //accumulate
#define endl "\n"
#define fi first
#define se second
#define FOR(i,s,t) for(int i=(s);i<=(t);++i)
#define mem(a,b) memset(a,b,sizeof(a))
#define debug(x) printf("%d\n",x)
using namespace std;
const int maxn=505;
long double a[maxn][maxn];
int c[maxn];
int n,m;
const long double eps=1e-8;
struct rec
{
    int i;
    bool operator<(const rec& rhs)const
    {
        return c[i]<c[rhs.i];
    }
};
set<rec> S;
int main()
{

    //cin.tie(0);
    //cout.tie(0);
    //ios_base::sync_with_stdio(false);
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    scanf("%d%d",&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++)
    {
        scanf("%d",&c[i]);
    }
    int dim=0,ans=0,zh=min(n,m);
    for(int i=1; i<=zh; i++)
    {
        S.clear();
        for(int j=dim+1; j<=n; j++)
        {
            if(fabs(a[j][i])>eps)
            {
                S.insert((rec){j});
            }
        }
        if(S.empty())
        {
            continue;
        }
        dim++;
        int j=S.begin()->i;
        ans+=c[j];
        for(int k=dim; k<=m; k++)
        {
            swap(a[dim][k],a[j][k]);
        }
        swap(c[dim],c[j]);
        for(j=1; j<=n; j++)
        {
            if(i==j||fabs(a[j][i])<eps)
                continue;
            long double rate=a[j][i]/a[i][i];
            for(int k=i; k<=m; k++)
            {
                a[j][k]-=rate*a[dim][k];
            }
        }
    }
    printf("%d %d",dim,ans);
}

/*
void read()
{

    char c = getchar();
    int x = 0;
    for (; (c < 48 || c>57); c = getchar());
    for (; c > 47 && c < 58; c = getchar())
    {
        x = (x << 1) + (x << 3) + c - 48;
    }
    return x;
}
*/