UVa 1312

很有成就感的一道题，做的时候睡觉前都在想。。好不容易想出来了，第二天早上起来忘了。。
本题结构和例题“最大子矩阵很像”，但用这个做法是不可能的，因为“场地”太大了。由于只有100个点，所以应该要枚举这100个点
读入之后先将x和y排序并去重
枚举 x和y,不是表示点，而是两条“边界”，将整个图形分成4个部分，然后依次按照四个方向二分计算最大的边长
时间： 10000个点的枚举，每个都需要4*log(10000)所以没问题

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <ctime>#include <cmath>#include <vector>#include <deque>#include <algorithm>#include <list>#include <map>#include <set>#include <queue>#include <stack>#define REP(i,n) for(int i=0;i<(n);i++)#define REP1(i,n) for(int i=1;i<=(n);i++)#define FOR(i,a,b) for (int i=(a);i<=(b);i++)#define CLR(x,n) memset(x,n,sizeof(x))#define PN printf("\n")using namespace std;int n,w,h,lx,ly,x[110],y[110],p[110][2],ansx,ansy,ans;void init(){        scanf("%d%d%d",&n,&w,&h);        x[0]=0;x[1]=w;y[0]=0;y[1]=h;        REP(i,n){                 scanf("%d%d",&x[i+2],&y[i+2]);                 p[i][0]=x[i+2];p[i][1]=y[i+2];        }        lx=n+2;ly=n+2;        sort(x,x+lx);sort(y,y+ly);        lx=unique(x,x+lx)-x;ly=unique(y,y+ly)-y;        //REP(i,lx)printf("%d ",x[i]);PN;REP(i,ly)printf("%d ",y[i]);PN;}bool test4(int x1,int x2,int y1,int y2){     REP(i,n){              int x0=p[i][0],y0=p[i][1];              if ( (x0<x2 && x0>x1) && (y0<y2 && y0>y1) ) return false;     }     return true;}bool test2(int x0,int y0,int dir,int len){     //printf("  t2 %d %d %d %d\n",x0,y0,dir,len);     if (dir==1) return test4(x0,x0+len,y0,y0+len);     if (dir==2) return test4(x0-len,x0,y0,y0+len);     if (dir==3) return test4(x0-len,x0,y0-len,y0);     if (dir==4) return test4(x0,x0+len,y0-len,y0);}int test1(int x0,int y0,int dir,int mx){    //printf(" t1 %d %d %d %d\n",x0,y0,dir,mx);    if (mx<=ans)return 0;    int l=0,r=mx,mid;    while (l<r)    {          mid=l+(r-l+1)/2;          if (test2(x0,y0,dir,mid))l=mid;          else r=mid-1;    }    //printf(" Result: %d\n",l);    return l;}void renew(int& ans0,int ned,int &dir0,int dir1){ if (ned>ans0) {ans0=ned;dir0=dir1;}}int test(int x0,int y0,int &dir){    //printf("t0 %d %d\n",x0,y0);    int mx,ans0=0;    mx=min(x0,y0);renew(ans0,test1(x0,y0,3,mx),dir,3);    mx=min(w-x0,h-y0);renew(ans0,test1(x0,y0,1,mx),dir,1);    mx=min(x0,h-y0);renew(ans0,test1(x0,y0,2,mx),dir,2);    mx=min(w-x0,y0);renew(ans0,test1(x0,y0,4,mx),dir,4);    //printf("Result %d\n",ans0);    return ans0;}void make1(int x0,int y0,int dir,int &ax,int &ay,int te){     if (dir==1) {ax=x0;ay=y0;}     if (dir==2) {ax=x0-te;ay=y0;}     if (dir==3) {ax=x0-te;ay=y0-te;}     if (dir==4) {ax=x0;ay=y0-te;}}void solve(){     ans=0;int te,dir;     REP(i,lx)REP(j,ly)     {                       //printf("%d %d\n",x[i],y[j]);                       te=test(x[i],y[j],dir);                       if (te>ans)  {                                    make1(x[i],y[j],dir,ansx,ansy,te);                                    ans=te;                       }     }     printf("%d %d %d\n",ansx,ansy,ans);}int main(){    int ka;scanf("%d",&ka);    while (ka--)    {          init();          solve();          if (ka)PN;    }     system("pause");    return 0;}