Bzoj3007: Save the Little Cloud Princess

Descriptionthe hero is about to embark on the path of saving the Princess ...this time the goal of salvation is-love and justice of the Princess Xiao Yun. the hero came to the door of the boss's cave, he suddenly was ignorant, because in front of not just a boss, but on thousands boss. When the hero realizes he is still level 1, he understands that this is an impossible task. But he did not give up, he was thinking, can avoid the boss to save the princess, hehe. The boss's Cave can be viewed as a rectangle, the hero in the lower-left corner (Row,line), and the princess in the upper-right corner. Heroes in order to avoid bosses, of course, the farther away from the boss, the better, so the hero decided to find a path to the shortest distance from the boss farthest. Ps: The hero walks the direction is arbitrary. can you help him? when the hero found a beautiful princess Xiao Yun, immediately surrounded by the Boss!!! The hero slowly closed his eyes, hand flick, white light after the use of the back of the town scroll, returned to the castle, but only Xiao Yun Princess back ... Because the hero forgot to enter the town of the law. InputThe first line, enter three integers, n indicates the number of bosses, Row,line represents the size of the rectangle;next n rows, two integers per line represent the position coordinates of the boss. Outputoutputs a decimal number that indicates the hero's path is farthest from the boss and is accurate to two digits after the decimal point. The two-part answer R, the circle of radius R for each point, determines whether the circle is cut off (n,m), that is, from the left and the upper boundary can go through the circle to the right/lower boundary

#include <cstdio>#include<cmath>#include<algorithm>intN;DoubleXm,ym;structpos{Doublex, y;} ps[3010];intq[3010],ed[3010];BOOLDISCMP (DoubleXDoubleYDoublez) {    returnx*x+y*y<z*Z;}BOOLChkDoubleR) {    intQl=0, qr=0;  for(intI=0; i<n;i++){        if(ps[i].x-1<r| | Ps[i].y+r>ym) {Q[QR++]=i; Ed[i]=1; if(ps[i].x+r>xm| | ps[i].y-r<1)return 1; }Elseed[i]=0; }     while(ql!=qr) {        intw=q[ql++];  for(intI=0; i<n;i++)if(!ED[I]&AMP;&AMP;DISCMP (ps[i].x-ps[w].x,ps[i].y-ps[w].y,r*2) ) {Q[QR++]=i; Ed[i]=1; if(ps[i].x+r>xm| | ps[i].y-r<1)return 1; }    }    return 0;}intMain () {scanf ("%D%LF%LF",&n,&xm,&ym);  for(intI=0; i<n;i++) {scanf ("%LF%LF",&ps[i].x,&ps[i].y); }    DoubleL=0, r=xm+ym+1;  while(r-l>1e-5){        DoubleM= (L+R)/2.; if(CHK (M)) R=M; ElseL=l; } printf ("%.2lf\n", (L+R)/2.); return 0;}

You can also use the Kruskal of the smallest spanning tree to add edges until the first edge is found to connect the left and top borders to the right and bottom boundaries

#include <cstdio>#include<cmath>#include<algorithm>intn,xm,ym,ep=0;intf[3010];structpos{intx, y;} ps[3010];structedge{intb;Long LongC;} e[10000010];BOOL operator<(Edge A,edge b) {returna.c<B.C;} InlineLong LongMin2 (Long LongALong Longb) {a*=a;b*=b; returnA<b?a:b;}int Get(intx) {    intA=X,c;  while(X!=f[x]) x=F[x];  while(x!= (C=f[a])) f[a]=x,a=C; returnx;}intMain () {scanf ("%d%d%d",&n,&xm,&ym);  for(intI=0; i<n;i++) {F[i]=i; scanf ("%d%d",&ps[i].x,&ps[i].y); E[ep++]= (Edge) {i,n,4*min2 (xm-ps[i].x,ps[i].y-1)}; E[ep++]= (Edge) {i,n+1,4*min2 (ps[i].x-1, ym-ps[i].y)};  for(intj=0; j<i;j++){            Long Longx=ps[i].x-ps[j].x,y=ps[i].y-ps[j].y; E[ep++]= (Edge) {i,j,x*x+y*y}; }} F[n]=n;f[n+1]=n+1; Std::sort (E,e+EP);  for(intI=0; i<ep;i++){                intA=Get(E[I].A), b=Get(E[I].B); if(a!=b) {F[a]=b; if(Get(n) = =Get(n+1) {printf ("%.2f\n", sqrt (E[I].C)/2. +1e-8); return 0; }        }    }    return 0;}

Bzoj3007: Save the Little Cloud Princess