POJ 2653 Pick-up sticks（几何）

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发布时间：2019-06-22

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Pick-up sticks

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 13377		Accepted: 5039

Description

Stan has n sticks of various length. He throws them one at a time on the floor in a random way. After finishing throwing, Stan tries to find the top sticks, that is these sticks such that there is no stick on top of them. Stan has noticed that the last thrown stick is always on top but he wants to know all the sticks that are on top. Stan sticks are very, very thin such that their thickness can be neglected.

Input

Input consists of a number of cases. The data for each case start with 1 <= n <= 100000, the number of sticks for this case. The following n lines contain four numbers each, these numbers are the planar coordinates of the endpoints of one stick. The sticks are listed in the order in which Stan has thrown them. You may assume that there are no more than 1000 top sticks. The input is ended by the case with n=0. This case should not be processed.

Output

For each input case, print one line of output listing the top sticks in the format given in the sample. The top sticks should be listed in order in which they were thrown.

The picture to the right below illustrates the first case from input.

Sample Input

51 1 4 22 3 3 11 -2.0 8 41 4 8 23 3 6 -2.030 0 1 11 0 2 12 0 3 10

Sample Output

Top sticks: 2, 4, 5.Top sticks: 1, 2, 3.

Hint

Huge input,scanf is recommended.

Source

题解：

　　几何模板题目

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 #include 
         
           5 #include 
          
            6 #include 
           
             7 #include 
            
              8 #include 
             
               9 #include 
              10 #include 
               
                11 #include 
                
                 12 using namespace std;13 typedef long long LL;14 #define ms(a, b) memset(a, b, sizeof(a))15 #define pb push_back16 #define mp make_pair17 #define eps 0.000118 const LL INF = 0x7fffffff;19 const int inf = 0x3f3f3f3f;20 const int mod = 1e9+7;21 const int maxn = 100000+10;22 int n, cnt;23 struct POINT24 {25 double x, y, z;26 POINT():x(0), y(0), z(0){};27 POINT(double _x_, double _y_, double _z_ = 0):x(_x_), y(_y_), z(_z_) {};28 };29 struct SEG30 {31 POINT a;32 POINT b;33 SEG(){};34 SEG(POINT _a_, POINT _b_):a(_a_),b(_b_) {};35 };36 double Cross(const POINT &a, const POINT & b, const POINT &o)//叉乘37 {38 return (a.x - o.x)*(b.y - o.y) - (b.x - o.x)*(a.y - o.y);39 }40 bool IsIntersect(const SEG &u, const SEG &v)41 {42 return (Cross(v.a, u.b, u.a)*Cross(u.b, v.b, u.a)>=0)&&43 (Cross(u.a, v.b, v.a)*Cross(v.b, u.b, v.a)>=0)&&44 (max(u.a.x, u.b.x) >= min(v.a.x, v.b.x))&&45 (max(v.a.x, v.b.x) >= min(u.a.x, u.b.x))&&46 (max(u.a.y, u.b.y) >= min(v.a.y, v.b.y))&&47 (max(v.a.y, v.b.y) >= min(u.a.y, u.b.y));48 }49 int ans[maxn];50 SEG stick[maxn];51 void init()52 {53 cnt = 0;54 }55 void solve() {56 double a, b, c, d;57 for(int i = 1;i<=n;i++){58 scanf("%lf%lf%lf%lf", &a, &b, &c, &d);59 stick[i].a.x=a;stick[i].a.y=b;60 stick[i].b.x=c;stick[i].b.y=d;61 }62 for(int i = n;i>0;i--){63 bool flag = 1;64 for(int j = i+1;j<=n;j++){65 if(IsIntersect(stick[i], stick[j])){66 flag = 0;67 break;68 }69 }70 if(flag){71 ans[cnt++] = i;72 }73 if(cnt>=1000){74 break;75 }76 }77 printf("Top sticks:");78 for(int i = cnt-1;i>=0;i--){79 if(i==0) printf(" %d.",ans[i]);80 else printf(" %d,",ans[i]);81 }82 printf("\n");83 }84 int main() {85 #ifdef LOCAL86 freopen("input.txt", "r", stdin);87 // freopen("output.txt", "w", stdout);88 #endif89 while(~scanf("%d", &n)&&n){90 init();91 solve();92 }93 return 0;94 }

View Code

转载于:https://www.cnblogs.com/denghaiquan/p/7223237.html

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