cogs103&tyvj1899 [NOIP2002] Rectangle Overlay

The scope of the topic is k<=4, but the official data is not k==4, resulting in some weird DP writing can also be passed. I heard that the calibration process in the k==4 when there are counter-examples, lift table ..... No wonder the answer to K==4 's data on cogs is wrong.

Well, write a search: There are a lot of writing on the Web. I'm going to split the set of points into two parts along a straight line parallel to the axis, and enumerate how K-rectangles are allocated on both sides. The boundary is k==1, sweep over all points to find the smallest rectangle. See the code for details.

But I cannot guarantee that this search is right because k==4 is likely to have this rugged and optimal solution: there is no line parallel to the axis and does not intersect any of the rectangles to divide the 4 rectangles into two parts. For example, the optimal scheme is:

Put a code: recursion in order to deal with "the point set divided into two parts" tuned a bunch of memcpy .... Fortunately, the number of recursive layers and points are not many, or the constant bombing days ....

#include <cstdio>#include<cstring>#include<algorithm>using namespacestd;Const intmaxn= -;intX[maxn],y[maxn],seq[maxn];inlineintMaxintAintb) {    returnA>b?a:b;} InlineintMinintAintb) {    returnA<b?a:b;}BOOLCMP1 (Const int&a,Const int&b) {    returnx[a]<x[b];}BOOLCMP2 (Const int&a,Const int&b) {    returny[a]<y[b];}structnode{intX1,y1,x2,y2; Node () {} node (intAintBintCintd) {X1=a;y1=b;x2=c;y2=D; }}sol[Ten];intCnt=0;intWorkintLintRintseq[]) {    intminx=0x7f7f7f7f, miny=0x7f7f7f7f, maxx=0, maxy=0;  for(intI=l;i<=r;++i) {//printf ("seq%d\n", Seq[i]);Minx=min (Minx,x[seq[i]); maxx=Max (Maxx,x[seq[i]]); Miny=min (Miny,y[seq[i]); maxy=Max (Maxy,y[seq[i]]); } sol[++cnt]=node (minx,miny,maxx,maxy); return(Maxx-minx) * (maxy-miny);}ints[ +][ -];inttot=0;intDfsintLintRintKintSeq[]) {//printf ("%d\n", k);    if(k==1){        inttmp=Work (L,R,SEQ); CNT--; returntmp; }Else{        intans=0x7f7f7f7f; Sort (Seq+l,seq+r+1, CMP1);  for(inti=l;i<r;++i) {             for(intj=1; j<k;++j) {                ++tot; memcpy (S[tot],seq,sizeof(int)* -); ++tot; memcpy (S[tot],seq,sizeof(int)* -); if(x[seq[i]]!=x[seq[i+1]]){                    intTmp=dfs (l,i,j,s[tot-1]) +dfs (i+1, r,k-J,s[tot]); Ans=min (ans,tmp); }                --tot;--tot; //printf ("---------------\ n");            }        }//printf ("%d\n", ans);Sort (seq+l,seq+r+1, CMP2);  for(inti=l;i<r;++i) {             for(intj=1; j<k;++j) {                ++tot; memcpy (S[tot],seq,sizeof(int)* -); ++tot; memcpy (S[tot],seq,sizeof(int)* -); if(y[seq[i]]!=y[seq[i+1]]){                    intTmp=dfs (l,i,j,s[tot-1]) +dfs (i+1, r,k-J,s[tot]); Ans=min (ans,tmp); }                --tot;--tot; //printf ("---------------\ n");            }        }//printf ("%d\n", ans);        returnans; }}intMain () {intn,k; scanf ("%d%d",&n,&k);  for(intI=1; i<=n;++i) {scanf ("%d%d", x+i,y+i); }     for(intI=1; i<=n;++i) {Seq[i]=i; } printf ("%d\n", DFS (1, N,k,seq)); return 0;}

cogs103&tyvj1899 [NOIP2002] Rectangle Overlay