bzoj3550 [ontak2010]vacation Simple Form __

Source: Internet
Author: User

Description

There are 3N number, you need to select some number, first of all, to ensure that any length of the interval of n, the number of selected <=k, and secondly to ensure that the number of the most selected.

Input

The first line is a two integer n,k.
The second row has a 3N integer.

Output

One-line integer represents the answer.

Sample Input

5 3

14 21 9 30, 11 8 1 20 29 23 17 27 7 8 35

Sample Output

195

HINT

"Data Range"

n<=200,k<=10.

Exercises
Bare-shaped problem.
Constructed according to the conditions.

Code

#include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #include <
cmath> #include <set> #include <complex> #define LL long #define MOD 10000007 using namespace std;
    inline int read () {int X=0,f=1;char ch=getchar (); while (ch< ' 0 ' | |
    Ch> ' 9 ') {if (ch== '-') F=-1;ch=getchar ();}
    while (ch>= ' 0 ' &&ch<= ' 9 ') {x=x*10+ch-' 0 '; Ch=getchar ();}
return x*f;
#define EPS (1e-10) const int n=1005;
int n,m,k;
Double A[n][n],b[n],c[n],ans;
    inline void pivot (int l,int e) {b[l]/=a[l][e];
    for (int i=1;i<=n;i++) if (i!=e) a[l][i]/=a[l][e];
    A[l][e]=1.0/a[l][e];
        for (int i=1;i<=m;i++) if (I!=l&&fabs (a[i][e)) >eps) {b[i]-=a[i][e]*b[l];
        for (int j=1;j<=n;j++) if (j!=e) a[i][j]-=a[i][e]*a[l][j];
    A[i][e]=-a[i][e]*a[l][e];
    } Ans+=c[e]*b[l];
    for (int i=1;i<=n;i++) if (i!=e) c[i]-=c[e]*a[l][i];
C[e]=-c[e]*a[l][e];
  } void simplex () {int l,e;  while (1) {for (e=1;e<=n;e++) if (c[e]>eps) a break;
        if (e==n+1) break;
        Double t=10000000000;
            for (int i=1;i<=m;i++) if (a[i][e]>eps&&b[i]/a[i][e]<t) {t=b[i]/a[i][e];
        L=i;
    } pivot (l,e);
    int main () {scanf ("%d%d", &n,&k);
    for (int i=1;i<=3*n;++i) scanf ("%lf", &c[i));
        for (int i=1;i<=2*n+1;++i) {b[i]=k+0.0;
    for (int j=i;j<=i+n-1;++j) a[i][j]=1.0;
    for (int i=2*n+2;i<=5*n+1;++i) b[i]=1,a[i][i-2*n-1]=1;
    m=5*n+1;n*=3;
    Simplex ();
    printf ("%.0LF", ans+eps);
return 0; }

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