A brain puzzle worth writing!
At first, there was no thought, and it was only after reading the solution.
Considering the and representation of A and b as coordinates, then the point that may be the answer must be the point in the lower convex packet, so that the optimization enumeration guarantees efficiency.
However, it is not known exactly which points are the points in the convex package, but we know that the boundary of two points (a minimum of B minimum) must be on the convex hull, if it is known that the points on the two convex hull, we connect them, then the point of the original image is the farthest from this line, it must be the point on the convex hull, so it can be two points, the cross product of the vector to find out
1#include <iostream>2#include <cstring>3#include <cstdio>4 using namespacestd;5 Const intmaxn= -;6 Const intinf=1147483647;7 intA[MAXN][MAXN],B[MAXN][MAXN];8 intW[MAXN][MAXN],SX[MAXN],SY[MAXN];9 intMAT[MAXN],SLA[MAXN];Ten intLX[MAXN],LY[MAXN]; One intn,t; A structdot{ - intx, y; -Dot (intx_=0,inty_=0): X (x_), Y (y_) {} theFriendBOOL operator==(Dot A,dot b) { - returna.x==b.x&&a.y==b.y; - } - }lo,hi; + BOOLDFS (intx) { -sx[x]=1; + for(inty=1; y<=n;y++) A if(!Sy[y]) { at intt=lx[x]+ly[y]-W[x][y]; - if(t) sla[y]=min (sla[y],t); - Else{ -sy[y]=1; - if(!mat[y]| |DFS (Mat[y])) -{mat[y]=x;return true;} in } - } to return false; + } - the Dot KM () { *memset (LX,0,sizeof(LX)); $memset (Ly,0,sizeof(ly));Panax NotoginsengMemset (Mat,0,sizeof(MAT)); - for(intI=1; i<=n;i++) the for(intj=1; j<=n;j++) +lx[i]=Max (lx[i],w[i][j]); A the for(intx=1; x<=n;x++){ +memset (SLA, the,sizeof(SLA)); - while(true){ $memset (SX,0,sizeof(SX)); $memset (SY,0,sizeof(SY)); - if(DFS (x)) Break; - intmin=INF; the for(intI=1; i<=n;i++)if(!sy[i]) min=min (min,sla[i]); - for(intI=1; i<=n;i++){Wuyi if(Sx[i]) lx[i]-=Min; thesy[i]?ly[i]+=min:sla[i]-=Min; - } Wu } - } AboutDOT ret (0,0); $ for(intI=1; i<=n;i++){ -ret.x+=A[mat[i]][i]; -ret.y+=B[mat[i]][i]; - } A returnret; + } the - intSolve (Dot L,dot r) { $ for(intI=1; i<=n;i++) the for(intj=1; j<=n;j++) thew[i][j]=a[i][j]* (R.Y-L.Y)-b[i][j]* (r.x-l.x); theDot mid=KM (); the if(mid==l| | MID==R)returnMin (l.x*l.y,r.x*r.y); - returnmin (Solve (l,mid), Solve (Mid,r)); in } the the About intMain () { theFreopen ("frame.in","R", stdin); theFreopen ("Frame.out","W", stdout); thescanf"%d",&T); + while(t--){ -scanf"%d",&n); the for(intI=1; i<=n;i++) for(intj=1; j<=n;j++) scanf ("%d",&A[i][j]); Bayi for(intI=1; i<=n;i++) for(intj=1; j<=n;j++) scanf ("%d",&B[i][j]); the for(intI=1; i<=n;i++) for(intj=1; j<=n;j++) w[i][j]=-a[i][j];lo=KM (); the for(intI=1; i<=n;i++) for(intj=1; j<=n;j++) w[i][j]=-b[i][j];hi=KM (); -printf"%d\n", Solve (Lo,hi)); - } the return 0; the}
Graph theory (km algorithm, Brain puzzle): Hnoi 2014 Frame (frame)