Surround the Trees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 10464    Accepted Submission(s): 4062 Problem Description There are 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 are 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 are no more than 100 trees.   Input The input contains one or more data sets. At first line of each input data set is number of trees in this data set, it is followed by series of coordinates of the trees. Each coordinate is a positive integer pair, and each integer is less than 32767. Each pair is separated by blank. Zero at line for number of trees terminates the input for your program.   Output 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

分析：给n个点，求出把所有点包围进去的所有边的和。

计算凸包的步骤：

平面上三个点:p1(x1,y1),p2(x2,y2),p3(x3,y3)

s(p1,p2,p3)=(x1-x3)*(y2-y3)-(x2-x3)*(y1-y3)

如果s>0 则说明 这连接这3个点时是按照逆时针的顺序,如果是s<0则说明连接这3个点是按照顺时针的顺序

1、排序：先按y从小到大排序，y相等时，按x从下到大排序。

bool cmp(node a,node b) { if(a.y==b.y) return a.x<b.x; return a.y<b.y; } 2、判断n是否大于3（小于3围不成凸多边形），然后对初始三个点赋值。

if(n==1) return 0; if(n==2) return dist(ans[0],ans[1]); res[0]=ans[0]; res[1]=ans[1]; res[2]=ans[2]; 3、逆时针方向计算，先从右半边开始算，从第2个点开始，如果新加入的点，与之前确定点的向量的叉乘小于0的话，说明该点在之前的点的右面，不满足凸包，所以删除之前确定的点，top--

for(i=2;i<n;i++){ while(top && mult(ans[i],res[top],res[top-1])) //左转退栈，上凸包 top--; res[++top]=ans[i]; }

4、再从左半边开始算，从第2个点开始，如果新加入的点，与之前确定点的向量的叉乘小于0的话，说明该点在之前的点的左，不满足凸包，所以删除之前确定的点，top--

for(i=n-3;i>=0;i--){ while(top!=len && mult(pnt[i],res[top],res[top-1]))//下凸包。 top--; res[++top]=pnt[i]; } 5、向量叉乘函数

bool mult(node a,node b,node t) { return (a.x-t.x)*(b.y-t.y)>=(b.x-t.x)*(a.y-t.y); } AC代码如下：

#include <stdio.h> #include <algorithm> #include <math.h> using namespace std; struct node{ int x,y; }ans[105]; bool cmp(node a,node b) { if(a.y==b.y) return a.x<b.x; return a.y<b.y; } double dist(node a,node b) { return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y)); } bool mult(node a,node b,node t) { return (a.x-t.x)*(b.y-t.y)>=(b.x-t.x)*(a.y-t.y); } double grahum(int n) { int len,i,j,k,top=1; node res[105]; sort(ans,ans+n,cmp); if(n==1) return 0; if(n==2) return dist(ans[0],ans[1]); res[0]=ans[0]; res[1]=ans[1]; res[2]=ans[2]; for(i=2;i<n;i++){ while(top && mult(ans[i],res[top],res[top-1])) //左转退栈，上凸包 top--; res[++top]=ans[i]; } len = top; res[++top]=ans[n-2]; for(i=n-3;i>=0;i--){ while(top!=len && mult(ans[i],res[top],res[top-1]))//下凸包。 top--; res[++top]=ans[i]; } double result=.0; for(i=1;i<=top;i++) { result+=dist(res[i],res[(i+1)>top?1:(i+1)]); } return result; } int main() { int n; int i,j; while(scanf("%d",&n),n) { for(i=0;i<n;i++) scanf("%d %d",&ans[i].x,&ans[i].y); printf("%.2lf\n",grahum(n)); } return 0; }