NYOJ 78 圈水池 (入门级凸包)

题目链接:nyoj 78  单调链凸包小结

题目讲解:本题考查的主要是凸包的用法,算是入门级的吧,当然前提是你接触过,平面几何:

AC代码:

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<vector>
 6 using namespace std;
 7 struct T
 8 {
 9     int x,y;
10     friend int operator < (T a, T b)
11 {
12     if((a.x<b.x) || (a.x==b.x && a.y<b.y))
13         return 1;
14     return 0;
15 }
16 }a[105],ans[105];
17 //用来判断第三点在当前两点构成的直线的左侧还是右侧,右侧返回值小于0;
18 double judge(T a, T b,T c)
19 {
20   return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);
21 }
22 int main()
23 {
24   int T,m;
25   scanf("%d",&T);
26   while(T--)
27   {
28 
29       scanf("%d",&m);
30       for(int i=0; i<m; i++)
31       {
32         scanf("%d %d",&a[i].x,&a[i].y);
33       }
34    sort(a,a+m);
35    int k1=0;
36    for(int i=0; i<m; i++)//下凸包,从下面扫描个点;
37    {
38        while(k1>1 && judge(ans[k1-2],ans[k1-1],a[i])<=0)
39        {
40             k1--;
41        }
42      ans[k1++]=a[i];
43    }
44     int k2=k1;
45     for(int i=m-1; i>=0; i--)//上凸包,从上面扫描各点;
46    {
47        while(k1>k2 && judge(ans[k1-2],ans[k1-1],a[i])<=0)
48        {
49             k1--;
50        }
51      ans[k1++]=a[i];
52    }
53    k1--;
54    sort(ans,ans+k1);
55    for(int i=0; i<k1; i++)
56     printf("%d %d\n",ans[i].x,ans[i].y);
57   }
58   return 0;
59 }

 

posted on 2014-09-23 16:37  细雨微光  阅读(766)  评论(0编辑  收藏  举报