Surround the Trees (convex bag)

Surround the Trees

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)

Total submission (s): 291 Accepted Submission (s): 140

Problem Descriptionthere is a lot of trees in an area. A peasant wants to buy a rope to surround all these trees. So at first he must know the minimal required length of the rope. However, he does not know how to calculate it. Can you help him? The diameter and length of the trees is omitted, which means a tree can be seen as a point. The thickness of the rope is also omitted which means a rope can be seen as a line. There is no more than trees.

Input the input contains one or more data sets. At first line of each input data set are number of trees in this data set, it is followed by series of coordinates of the T Rees. Each coordinate is a positive an integer pair, and each of the integers is less than 32767. Each pair are separated by blank. Zero at line for number of trees terminates the input for your program.

Output of the minimal length of the rope. The precision should be 10^-2.

Sample Input 9 12 7 24 9 30 5 41 9 80 7 50 87 22 9 45 1 50 7 0

Sample Output 243.06

Sourceasia 1997, Shanghai (Mainland China)

Recommendignatius.l

/*
You need a special sentence when you're on. Especially when it comes to n==2, it is the distance between the returns, not twice times the distance.
*/
#include <bits/stdc++.h>using namespacestd;Const DoubleEPS = 1e-8;Const DoublePI = ACOs (-1.0);intSgnDoublex) {    if(Fabs (x) < EPS)return 0; if(X <0)return-1; Else return 1;}structpoint{Doublex, y; Point () {}, point (Double_x,Double_y) {x= _x;y =_y; } Pointoperator-(ConstPoint &b)Const    {        returnPoint (X-b.x,y-b.y); }    //Cross Product    Double operator^(ConstPoint &b)Const    {        returnx*b.y-y*b.x; }    //dot Product    Double operator*(ConstPoint &b)Const    {        returnx*b.x + y*b.y; }    voidinput () {scanf ("%LF%LF",&x,&y); }};structLine {point s,e; Line () {} line (point _s,point _e) {s= _s; E =_e; }}; //* Distance between two pointsDoubleDist (Point A,point b) {returnsqrt (A-B) * (Ab));} /** Find convex hull, Graham algorithm * point number 0~n-1* return convex hull result Stack[0~top-1] is the number of the convex hull*/Const intMAXN =1010; Point LIST[MAXN];intStack[maxn],top;//Polar Order relative to list[0]BOOL_cmp (Point p1,point p2) {DoubleTMP = (p1-list[0]) ^ (p2-list[0]); if(SGN (TMP) >0)        return true; Else if(SGN (tmp) = =0&& sgn (Dist (p1,list[0])-Dist (p2,list[0])) <=0)        return true; Else         return false;}voidGraham (intN)    {Point p0; intK =0; P0= list[0]; //find the bottom one .     for(inti =1; I < n;i++)    {        if((P0.y > list[i].y) | | (P0.y = = List[i].y && p0.x >list[i].x)) {P0=List[i]; K=i; }} swap (list[k],list[0]); Sort (List+1, list+n,_cmp); if(n = =1) {Top=1; stack[0] =0; return; }    if(n = =2) {Top=2; stack[0] =0; stack[1] =1; return ; } stack[0] =0; stack[1] =1; Top=2;  for(inti =2; I < n;i++)    {         while(Top >1&& SGN (list[stack[top-1]]-list[stack[top-2]]) ^ (list[i]-list[stack[top-2]]) <=0) Top--; Stack[top++] =i; }}   intN;intMain () {//freopen ("In.txt", "R", stdin);     while(SCANF ("%d", &n)!=eof&&N) {         for(intI=0; i<n;i++) {list[i].input (); }        if(n==1) {printf ("0.00\n"); Continue; }Else if(n==2) {printf ("%.2lf\n", Dist (list[0],list[1])); Continue;        } Graham (n); DoubleCur=0;  for(intI=0; i<top;i++) {cur+=dist (list[stack[i%top]],list[stack[(i+1)%top]]); } printf ("%.2lf\n", cur); }    return 0;} 

Surround the Trees (convex bag)