Rokua P1378 Oil Droplet expansion

Title Description

In a rectangular frame, a maximum of N (0≤n≤6) of different points, at any one of the points placed a very small oil droplets, then the oil droplets will continue to expand until exposed to other oil droplets or frame boundaries. You must wait for an oil droplet to expand before you can place the next drop of oil. Then, in what order should the oil droplets be placed on these n points, in order to make all the oil droplets occupy the overall product maximum.

Input Format:

Line 1th an integer n.
2nd Behavior Rectangle border The coordinates of a vertex and its diagonal vertices, x,y,x ', y '.
next n rows, two integers per line, xi,yi, representing the coordinates of the box's n points.
all of the above data is within [ -1000,1000].

output Format:

A row, an integer, the smallest space remaining in a rectangular box (results rounded out)

Input Sample # #:

2
0
3
17 7

Sample # # of output:

50

Ideas

Preprocess out points with the edges and the rest of the distance, DFS, note that only the oil droplets that have been placed will have an effect.

#include <cstdio> #include <cmath> using namespace std;
int n,x,y,xx,yy,a[100],b[100];
Double F[100][100],max;
BOOL t[100];
    Double min (double v,double e) {if (v<e) return v;
else return e;
    } int abs (int v) {if (v<0) return-v;
else return v;
        } int Sqr (int c) {return c*c;} int ss (int c,double p) {if (c>n) {if (P>max) max=p;
    return 0;
            } for (int i=1;i<=n;i++) {if (T[i]) {t[i]=false;
            Double o=0;
            O=min (ABS (A[I]-X), ABS (Xx-a[i));
            O=min (O,abs (b[i]-y));
            O=min (O,abs (yy-b[i));
                for (int j=1;j<=n;j++) {if (f[i][j]<=0) o=min (f[i][j],o);
            if (not T[j]) o=min (f[i][j],o);
                } if (o<=0) {t[i]=true;
            SS (C+1,P);
             } else {   for (int j=1;j<=n;j++) {f[i][j]=f[i][j]-o;
                F[J][I]=F[I][J];
                } SS (C+1,p+3.1415926*o*o);
                T[i]=true;
                    for (int j=1;j<=n;j++) {f[i][j]=f[i][j]+o;
                F[J][I]=F[I][J];
    }}}}} int main () {scanf ("%d", &n);
    scanf ("%d%d%d%d", &x,&y,&xx,&yy);
    for (int i=1;i<=n;i++) for (int j=1;j<=n;j++) f[i][j]=0xfffffff;
        for (int i=1;i<=n;i++) {scanf ("%d%d", &a[i],&b[i]);
            for (int k=1;k<i;k++) {f[i][k]=sqrt (SQR (ABS (A[I]-A[K))) +SQR (ABS (B[I]-B[K)));
        F[K][I]=F[I][K];              
    }} for (int i=1;i<=n;i++) t[i]=true;
    SS (1,0);
    Max=abs (xx-x) *abs (yy-y)-max;  
printf ("%d", int (max+0.5)); }