zoukankan      html  css  js  c++  java
  • hdu1392(凸包)

    Surround the Trees

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 7356    Accepted Submission(s): 2813


    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
    题意:求凸包周长
     1 #include<iostream>
     2 #include<stdio.h>
     3 #include<math.h>
     4 #include<algorithm>
     5 using namespace std;
     6 #define N 110
     7 struct point
     8 {
     9     double x,y,angel;
    10 } p[N],stack[N];
    11 int top,n;
    12 
    13 double dis(point a,point b)
    14 {
    15     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
    16 }
    17 
    18 bool mult(point p1,point p2,point p0)
    19 {
    20     return (p1.x-p0.x)*(p2.y-p0.y) >= (p2.x-p0.x)*(p1.y-p0.y);
    21 }
    22 
    23 bool cmp(point a,point b)
    24 {
    25     if(a.angel == b.angel)
    26     {
    27         if (a.x == b.x)
    28             return a.y > b.y;
    29         return a.x > b.x;
    30     }
    31     return a.angel < b.angel;
    32 }
    33 
    34 void graham()
    35 {
    36     int i,k=0;
    37     for(i=0; i<n; i++)
    38         if(p[i].y<p[k].y||((p[i].y==p[k].y)&&(p[i].x<p[k].x)))
    39             k=i;
    40     swap(p[0],p[k]);
    41     for(i=1; i<n; i++)
    42         p[i].angel=atan2(p[i].y-p[0].y,p[i].x-p[0].x);
    43     sort(p+1,p+n,cmp);
    44     stack[0]=p[0];
    45     stack[1]=p[1];
    46     stack[2]=p[2];
    47     top=3;
    48     for(i=3; i<n; i++)
    49     {
    50         while(top > 2 && mult(stack[top-2],stack[top-1],p[i])<=0)
    51             top--;
    52         stack[top++]=p[i];
    53     }
    54 }
    55 
    56 int main()
    57 {
    58     int i;
    59     double ans;
    60     while(scanf("%d",&n)!=EOF&&n)
    61     {
    62         for(i=0; i<n; i++)
    63             scanf("%lf%lf",&p[i].x,&p[i].y);
    64         if(n==1)
    65         {
    66             printf("0.00
    ");
    67             continue;
    68         }
    69         if(n==2)
    70         {
    71             printf("%.2lf
    ",dis(p[0],p[1]));
    72             continue;
    73         }
    74         graham();
    75         ans=0;
    76         for(i=0; i<top-1; i++)
    77             ans+=dis(stack[i],stack[i+1]);
    78         ans+=dis(stack[top-1],stack[0]);
    79         printf("%.2lf
    ",ans);
    80     }
    81     return 0;
    82 }
    View Code
  • 相关阅读:
    P2567 [SCOI2010]幸运数字 DFS+容斥定理
    Codeforces Round #462 (Div. 2) C DP
    Codeforces Round #428 (Div. 2) C. dfs
    POJ 2079 最大三角形面积(凸包)
    POJ 3608 凸包间最短距离(旋转卡壳)
    2018年全国多校算法寒假训练
    Educational Codeforces Round 37 E. Connected Components?(图论)
    UVa 1440:Inspection(带下界的最小流)***
    BZOJ 1483:[HNOI2009]梦幻布丁(链表启发式合并)
    PAT L3-016:二叉搜索树的结构(暴力)
  • 原文地址:https://www.cnblogs.com/lxm940130740/p/3899636.html
Copyright © 2011-2022 走看看