Bzoj 1696 [Usaco2007 feb]building A new Barn Barn Math

Test Instructions:Link Method:Math + Simulation parsing:First of all, this kind of problem is not the first time to see, so directly know to take the median of x, y median. The problem is that it is very annoying to discuss the situation. There is a limitation in the question, given that the point to be summed cannot be selected. So if an odd number of points, find the x median, y median number. Verify that x, y exists in the set of points to be summed, and four cases are judged if present. The four cases of judgement are (x-1,y), (X+1,y), (x,y-1), (x,y+1), suboptimal solution must exist in these four cases, this should be very good understanding? If it does not exist in the origin set, it is summed directly. If you have an even number of points, it is advisable to discuss all the X-values and the Y-values, and we recommend using the repulsion. As for the details, study for yourself. Code:

#include <map>#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define N 10010#define INF 0x3f3f3f3f#define PA pair<int,int>using namespace STD;intNintA[n];intB[n];intxx[5]={0,-1,1,0,0};intyy[5]={0,0,0,1,-1};structnode{intx, y;} C[n];intMain () {scanf("%d", &n); for(intI=1; i<=n;i++) {scanf("%d%d", &a[i],&b[i]);    C[i].x=a[i],c[i].y=b[i]; } sort (A +1, a+n+1), sort (b +1, b+n+1);if(n&1)    {inttmp=n/2+1;intTMPX=A[TMP],TMPY=B[TMP];intans=0;intflag=1; for(intI=1; i<=n;i++) {if(TMPX==C[I].X&AMP;&AMP;TMPY==C[I].Y) {flag=0; Break;} }if(flag) { for(intI=1; i<=n;i++) {ans+=ABS(Tmpx-a[i]) +ABS(Tmpy-b[i]); }printf("%d 1\n", ans); }Else{Ans=inf;intCnt=0; for(intI=1; i<=4; i++) {intX=tmpx+xx[i],y=tmpy+yy[i];intret=0; for(intj=1; j<=n;j++) {ret+=ABS(x-c[j].x) +ABS(Y-C[J].Y); }if(Ret<ans) ans=ret,cnt=1;Else if(Ret==ans) cnt++; }printf("%d%d\n", ans,cnt); }    }Else{inttmpx1=a[n/2],tmpx2=a[n/2+1];inttmpy1=b[n/2],tmpy2=b[n/2+1];intret=0, Cnt= (tmpx2-tmpx1+1) * (tmpy2-tmpy1+1); for(intk=1; k<=n;k++) {if(c[k].x>=tmpx1&&c[k].x<=tmpx2&&c[k].y>=tmpy1&&c[k].y<=tmpy2) cnt--; ret+=ABS(tmpx1-c[k].x) +ABS(TMPY1-C[K].Y); }printf("%d%d\n", ret,cnt); }}

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Bzoj 1696 [Usaco2007 feb]building A new Barn Barn Math