Aquarium Tank（csu1634+几何+二分）Contest2087 - 湖南多校对抗赛（2015.05.24）-G

 Aquarium Tank

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 15  Solved: 4
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Description

You just bought an “artistic” aquarium tank that has an interesting shape, and you poured L litres of water into the tank. How high is the water in the tank?
When you look at this tank from one side, it has the shape of a convex polygon. This polygon has exactly two vertices on the table (y-coordinates are 0), and all other vertices have positive y-coordinates. There are also exactly two vertices with maximum y-coordinates, and water is poured into the opening between these two vertices. This aquarium tank has a depth of D centimetres. The tank is glued to the table, so no matter what shape it has, it keeps its position and does not tip over. All coordinates and lengths in this problem are given in centimetres. It should be noted that each cubic metre is equivalent to 1 000 litres.
An illustration showing the configuration of the tank of the first sample input is given below:

Input

The input consists of a single test case. The first line contains an integer N (4 ≤ N ≤ 100) giving the number of vertices in the polygon. he next line contains two integers D and L, where 1 ≤ D ≤ 1000 is he depth of the aquarium tank and 0  L  2 000 is the number of litres f water to pour into the tank. The next N lines each contains two integers, giving the (x, y) coordinates of the vertices of the convex polygon in counterclockwise order. The absolute values of x and y are at most 1 000. You may assume that the tank has a positive capacity, and you never pour more water than the tank can hold.

 

Output

Print the height of the water (in centimetres) in the aquarium tank on a line to 2 decimal places.

 

Sample Input

4
30 50
20 0
100 0
100 40
20 40

Sample Output

20.83

HINT

 

Source



 

题意：有一个横放是多边形的棱柱。问L升水，注入其中，容器的深度是多少。

思路：棱柱的体积等于底面积成高，so、、、我可以二分多边形的高度（用平行与X轴的直线求切割多边形）取下半部分是面积；

坑！比赛的时候，我以为第一组边一定是在x轴上的。。。o(︶︿︶)o 唉，结果是一定有一组边在x轴上，但不一定是第一组！。。。

 

转载请注明出处：寻找&星空の孩子 

题目链接：http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1634

 

 

比较乱，没整理！

  1 #include<cstdio>
  2 #include<cmath>
  3 #include<iostream>
  4 #define PI acos(-1.0)
  5 using namespace std;
  6  
  7 struct Point
  8 {
  9     double x,y;
 10     Point(double x=0,double y=0):x(x),y(y) {} 
 11  
 12 };
 13  
 14 double hmax;
 15 double D,L;
 16  
 17 typedef Point Vector;//Vector只是Point的别名
 18  
 19 //向量+向量=向量;    向量+点=点
 20 Vector operator + (Vector A,Vector B)
 21 {
 22     return Vector(A.x+B.x,A.y+B.y);
 23 }
 24  
 25 //点-点=向量
 26 Vector operator - (Point A,Point B)
 27 {
 28     return Vector(A.x-B.x,A.y-B.y);
 29 }
 30  
 31 //向量*数=向量
 32 Vector operator * (Vector A,double p)
 33 {
 34     return Vector(A.x*p,A.y*p);
 35 }
 36  
 37 //向量/数=向量
 38 Vector operator / (Vector A,double p)
 39 {
 40     return Vector(A.x/p,A.y/p);
 41 }
 42  
 43  
 44 //
 45 bool operator < (const Point& a,const Point& b)
 46 {
 47     return a.x<b.x||(a.x==b.x && a.y<b.y);
 48 }
 49  
 50 //
 51 const double eps = 1e-8;
 52 //三态函数
 53 int dcmp(double x)
 54 {
 55     if(fabs(x)<eps)return 0;
 56     else return x < 0 ? -1 : 1;
 57 }
 58 //相等
 59 bool operator == (const Point& a,const Point& b)
 60 {
 61     return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0;
 62 }
 63  
 67 double Dot(Vector A,Vector B)
 68 {
 69     return A.x*B.x+A.y*B.y;
 70 }
 71 double length(Vector A)
 72 {
 73     return sqrt(Dot(A,A));
 74 }
 75 double Angle(Vector A,Vector B)
 76 {
 77     return acos(Dot(A,B)/length(A)/length(B));
 78 }
 79  
 84 double Cross(Vector A,Vector B)
 85 {
 86     return A.x*B.y-B.x*A.y;
 87 }
 88 double Area2(Point A,Point B,Point C)
 89 {
 90     return Cross(B-A,C-A);
 91 }
 92  
 93 double PArea(Point *p,int n)
 94 {
 95     double area=0;
 96     for(int i=0; i<n; i++)
 97     {
 98         area+=Cross(p[i],p[(i+1)%n]);
 99   //      printf("area=%f\n",area);
100     }
101     return fabs(area/2);
102 }
103  
104 /*******两直线交点*******/
105 //调用前确保两条直线P+tv和Q+tv有唯一交点，当且仅当Cross(v,w)非0;
106 Point GetLineIntersection(Point P,Vector v,Point Q,Vector w)
107 {
108     Vector u=P-Q;
109   //  printf("P=(%f,%f),v=(%f,%f),Q=(%f,%f),w=(%f,%f)\n",P.x,P.y,v.x,v.y,Q.x,Q.y,w.x,w.y);
110  //   if(Cross(v,w))
111  //   {
112   //      double t=Cross(w,u)/Cross(v,w);//精度高的时候，考虑自定义分数类
113    //     return P+v*t;
114   //  }
115 //    else
116 //       return ;
117 //printf("t=   %lf\t%lf",Cross(w,u),Cross(v,w));
118     return P+v*Cross(w,u)/Cross(v,w);
119 }
120 //输入函数
121 Point read_point(Point &P)
122 {
123     scanf("%lf%lf",&P.x,&P.y);
124     hmax=max(hmax,P.y);
125     return P;
126 }
127 Point get_point(Point a, Point b, double y0)
128 {
129     if(fabs(a.x - b.x) < eps) return Point(a.x, y0);
130     double bi = (y0 - a.y) / (b.y - a.y);
131     return Point(a.x + bi * (b.x - a.x), a.y + bi * (b.y - a.y));
132 }
133 int main()
134 {
135     Point po[105],Q[105];
136     int T,n,q,i;
137  
138   //  freopen("debug//secret-tank-03.in","r",stdin);
139     while(scanf("%d",&n)!=EOF)
140     {
141         scanf("%lf%lf",&D,&L);
142         hmax=0;
143         for(i=0; i<n; i++)
144         {
145             read_point(po[i]);
146   //          printf("----[%lf]\n",po[i]);
147         }
148    //     po[n]=po[0];
149  //       po[n+1]=po[1];
150   //      Q[0]=po[0];
151   //      Q[1]=po[1];
152     //   printf("Q[%d]=%lf\n\n",1,po[1]);
153         double d=0,h=hmax;
154         while(h-d>eps)
155         {
156             q=0;
157             int per=n-1;
158             double m=(d+h)/2;
159         //    printf("m=%lf\n",m);
160         //    getchar();
161             Point M(0,m);
162             Vector w(1,0);
163             for(int i=0; i<n; i++)
164             {
165       //          if(i==0&&(m-po[i].y)>eps){Q[q++]=po[i];continue;}
166  
167  
168   //              else if((m-po[i].y)==eps)
169   //              {
170   //                  Q[q++]=po[i];
171   //              }
172                 if((m-po[i].y)*(m-po[per].y)<eps)
173                 {
174                  //   printf("------------------------------------------------------\n");
175   //                  Vector PP=po[i]-po[per];
176    //                 Q[q++]=GetLineIntersection(po[per],PP,M,w);
177                     Q[q++]=get_point(po[i],po[per],m);
178                 }
179                 if((m-po[i].y)>eps)
180                 {
181         //            if(i==1) Q[q++]=po[i-1];
182                     Q[q++]=po[i];
183                 }
184                 per=i;
185                // printf("Q[%d]=(%f,%f)\n",q-1,Q[q-1].x,Q[q-1].y);
186             }
187     //        Q[q]=Q[0];
188  //   if(m==20)
189    //        for(int i=0;i<q;i++)
190      //       {
191        //         printf("Q[%d]=(%f,%f)\n",i,Q[i].x,Q[i].y);
192          //   }
193             double area=PArea(Q,q);
194            // printf("L=%f\tV=%lf\tm=%lf\n",L*1000,area*D,m);
195             if(L*1000-area*D>eps) d=m;
196             else h=m;
197  
198         }
199         printf("%.2f\n",d);
200     }
201     return 0;
202 }
203  
204 /**************************************************************
205     Problem: 1634
206     User: aking2015
207     Language: C++
208     Result: Accepted
209     Time:0 ms
210     Memory:1496 kb
211 ****************************************************************/

 

 

 

好吧。发个整理过的。

  1 #include<cstdio>
  2 #include<cmath>
  3 #include<iostream>
  4 #define PI acos(-1.0)
  5 using namespace std;
  6  
  7 struct Point
  8 {
  9     double x,y;
 10     Point(double x=0,double y=0):x(x),y(y) {}
 11  
 12 };
 13  
 14 double hmax;
 15 double D,L;
 16  
 17 typedef Point Vector;
 18  
 19 Vector operator + (Vector A,Vector B)
 20 {
 21     return Vector(A.x+B.x,A.y+B.y);
 22 }
 23  
 24 Vector operator - (Point A,Point B)
 25 {
 26     return Vector(A.x-B.x,A.y-B.y);
 27 }
 28  
 29 Vector operator * (Vector A,double p)
 30 {
 31     return Vector(A.x*p,A.y*p);
 32 }
 33  
 34 Vector operator / (Vector A,double p)
 35 {
 36     return Vector(A.x/p,A.y/p);
 37 }
 38  
 39 bool operator < (const Point& a,const Point& b)
 40 {
 41     return a.x<b.x||(a.x==b.x && a.y<b.y);
 42 }
 43  
 44 const double eps = 1e-8;
 45  
 46 int dcmp(double x)
 47 {
 48     if(fabs(x)<eps)return 0;
 49     else return x < 0 ? -1 : 1;
 50 }
 51 bool operator == (const Point& a,const Point& b)
 52 {
 53     return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0;
 54 }
 55  
 56 double Dot(Vector A,Vector B)
 57 {
 58     return A.x*B.x+A.y*B.y;
 59 }
 60 double length(Vector A)
 61 {
 62     return sqrt(Dot(A,A));
 63 }
 64 double Angle(Vector A,Vector B)
 65 {
 66     return acos(Dot(A,B)/length(A)/length(B));
 67 }
 68  
 69 double Cross(Vector A,Vector B)
 70 {
 71     return A.x*B.y-B.x*A.y;
 72 }
 73 double Area2(Point A,Point B,Point C)
 74 {
 75     return Cross(B-A,C-A);
 76 }
 77  
 78 double PArea(Point *p,int n)
 79 {
 80     double area=0;
 81     for(int i=0; i<n; i++)
 82     {
 83         area+=Cross(p[i],p[(i+1)%n]);
 84     }
 85     return fabs(area/2);
 86 }
 87  
 88 Point GetLineIntersection(Point P,Vector v,Point Q,Vector w)
 89 {
 90     Vector u=P-Q;
 91     return P+v*Cross(w,u)/Cross(v,w);
 92 }
 93 Point read_point(Point &P)
 94 {
 95     scanf("%lf%lf",&P.x,&P.y);
 96     hmax=max(hmax,P.y);
 97     return P;
 98 }
 99 Point get_point(Point a, Point b, double y0)
100 {
101     if(fabs(a.x - b.x) < eps) return Point(a.x, y0);
102     double bi = (y0 - a.y) / (b.y - a.y);
103     return Point(a.x + bi * (b.x - a.x), a.y + bi * (b.y - a.y));
104 }
105 int main()
106 {
107     Point po[105],Q[105];
108     int T,n,q,i;
109     while(scanf("%d",&n)!=EOF)
110     {
111         scanf("%lf%lf",&D,&L);
112         hmax=0;
113         for(i=0; i<n; i++)
114         {
115             read_point(po[i]);
116         }
117         double d=0,h=hmax;
118         while(h-d>eps)
119         {
120             q=0;
121             int per=n-1;
122             double m=(d+h)/2;
123             Point M(0,m);
124             Vector w(1,0);
125             for(int i=0; i<n; i++)
126             {
127                 if((m-po[i].y)*(m-po[per].y)<eps)
128                 {
129                     Vector PP=po[i]-po[per];
130                     Q[q++]=GetLineIntersection(po[per],PP,M,w);
131                     //   Q[q++]=get_point(po[i],po[per],m);
132                 }
133                 if((m-po[i].y)>eps)
134                 {
135                     Q[q++]=po[i];
136                 }
137                 per=i;
138             }
139             double area=PArea(Q,q);
140             if(L*1000-area*D>eps) d=m;
141             else h=m;
142         }
143         printf("%.2f\n",d);
144     }
145     return 0;
146 }
147  
148 /**************************************************************
149     Problem: 1634
150     User: aking2015
151     Language: C++
152     Result: Accepted
153     Time:0 ms
154     Memory:1500 kb
155 ****************************************************************/

 

posted @ 2015-05-26 01:07  寻找&星空の孩子  阅读(694)  评论(0)  编辑  收藏  举报