Bzoj 3550 [ontak2010]vacation linear programming

link : Method: Linear programming parsing: This is a fairly simple linear programming of simplicity. First we set Xi x_i to indicate whether the number I is selected. it is clear x1+x2+...+xn<=k x_1+x_2+...+x_n x2+x3...+xn+1<=k x_2+x_3...+x_{n+1} . .. x2n+1+x2n+2+...+x3n<=k x_{2n+1}+x_{2n+2}+...+x_{3n} Second 0<=xi<=1 0 and we want to maximize ∑3ni=1ci∗xi \sum_{i=1}^{3n}c_i*x_i The above equation is obviously a standard linear programming. The simple solution is OK. How complex is it to solve these things? If you look at the code, or you understand the whole process. It is obviously O (nm) when the limit is eliminated. But when we enumerate the variables with coefficients greater than 0 in the target function, the worst enumeration is O (n). so it's the worst thing to reach O (n^2*m), and it seems a bit close to the network stream. so, which one is faster? Maybe you can try it someday. Code:

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define N 610 #defi
NE INF 0x7f7f7f7f using namespace std;
Double a[n<<1][n<<1];
int N,k,tot;
    int check () {for (int i=1;i<=3*n;i++) if (a[0][i]>0) return i;
return 0;
        } void Simplex () {while (int t=check ()) {double limit=inf;
        int choseline;
            for (int i=1;i<=tot;i++) {if (a[i][t]<=0) continue;
        else if (a[i][0]/a[i][t]<limit) {limit=a[i][0]/a[i][t];choseline=i}
        } if (Limit==inf) {a[0][0]=inf;break;}
        Double di=a[choseline][t];
            for (int i=0;i<=3*n;i++) {if (i==t) A[choseline][i]/=di;
        A[choseline][i]/=di; for (int i=0;i<=tot;i++) {if (i==choseline| |
            a[i][t]==0) continue;
            if (i==0) a[i][0]+=a[choseline][0]*a[i][t];
            else a[i][0]-=a[choseline][0]*a[i][t]; Double L=A[I][T];
                for (int j=1;j<=3*n;j++) {if (j==t) a[i][j]=-l*a[choseline][j];
            else A[i][j]-=l*a[choseline][j];
    int main () {scanf ("%d%d", &n,&k);
        for (int i=1;i<=2*n+1;i++) {int to=i+n-1;
        tot++;
        for (int j=i;j<=to;j++) a[tot][j]=1;
    A[tot][0]=k;
        for (int i=1;i<=3*n;i++) {tot++;
    A[tot][i]=1,a[tot][0]=1;
    for (int i=1;i<=3*n;i++) scanf ("%lf", &a[0][i));
    Simplex ();
printf ("%.0lf\n", a[0][0]); }