EOJ-1708//POJ3334

题意：

　　有一个连通器，由两个漏斗组成（关于漏斗的描述见描述）。

　　现向漏斗中注入一定量的水，问最终水的绝对位置（即y轴坐标）

思路：

　　总体来说分为3种情况。

　　1.两个漏斗可能同时装有水。

　　2.只可能a漏斗有水。

　　3.只可能b漏斗有水。

　　于是可以二分枚举y的坐标。

　　关键在于对于某个y坐标来说，要求出新的交点，再求面积。

 #include <stdio.h>
 #include <string.h>
 #include <stdlib.h>
 #include <math.h>
 #include <algorithm>
 #include <iostream>
 using namespace std;

 const int maxn = ;
 const double eps = 1e-;
 const double inf = 999999999.99;

 struct Point{
     double x,y;
 }a[ maxn ],b[ maxn ],res[ maxn ],amid,bmid;

 double xmult( Point a,Point b,Point c ){
     double ans = (a.x-c.x)*(b.y-c.y) - (a.y-c.y)*(b.x-c.x);
     return ans;
 }

 int cmp( Point a,Point b ){
     if( a.x!=b.x ) return a.x<b.x;
     else return a.y>b.y;
 }

 double area( Point pnt[],int n ){
     double ans = ;
     for( int i=;i<n-;i++ ){
         ans += xmult( pnt[],pnt[i],pnt[i+] );
     }
     return fabs( 0.5*ans );
 }

 int main(){
     //freopen("out.txt","w",stdout);
     int T;
     scanf("%d",&T);
     while( T-- ){
         double aim;
         double ansY = ;
         scanf("%lf",&aim);
         int n1,n2;
         scanf("%d",&n1);
         double ymax = inf;
         int flag1 = -;
         for( int i=;i<n1;i++ ){
             scanf("%lf%lf",&a[i].x,&a[i].y);
             if( ymax>a[i].y ){
                 ymax = a[i].y;
                 flag1 = i;
             }
         }
         amid = a[ flag1 ];
         scanf("%d",&n2);
         ymax = inf;
         int flag2 = -;
         for( int i=;i<n2;i++ ){
             scanf("%lf%lf",&b[i].x,&b[i].y);
             if( ymax>b[i].y ){
                 ymax = b[i].y;
                 flag2 = i;
             }
         }
         bmid = b[ flag2 ];
         //input
         double aYmin = min( a[].y,a[n1-].y );
         double bYmin = min( b[].y,b[n2-].y );
         //printf("aYmin = %lf bYmin = %lf\n",aYmin,bYmin);
         double abYmax = max( aYmin,bYmin );
         double abYmin = min( amid.y,bmid.y );
         double L ,R ;
         //printf("L = %lf , R = %lf \n",L,R);
         int special = -;
         if( aYmin<=bmid.y )//a is lower
         {
             special = ;
         }
         else if( bYmin<=amid.y )
         {
             special = ;
         }
         if( special==- ){
             L = abYmin;
             R = min( aYmin,bYmin );
             while( L<R ){
                 double mid = (L+R)/2.0;
                 double sumArea = ;
                 /*******solve b******/
                 //printf("mid = %lf\n",mid);
                 if( mid>bYmin ){
                     int cnt = ;
                     double newY = bYmin;
                     int f = -;
                     for( int i=;i<n2;i++ ){
                         if( b[i].y<=newY ){
                             res[ cnt++] = b[ i ];
                             f = i;
                         }
                         else break;
                     }
                     if( f==- ){}
                     else{
                         Point tmp;
                         tmp.y = newY;
                         tmp.x = (b[ f+ ].x-b[ f ].x)*(newY-b[f].y)/(b[f+].y-b[f].y) + b[f].x;
                         res[ cnt++ ] = tmp;
                     }
                     sumArea += area( res,cnt );
                 }
                 else if( mid<=bmid.y ){}
                 else{
                     //printf("here\n");
                     int cnt = ;
                     int f = -;
                     for( int i=;i<n2;i++ ){
                         if( b[i].y<=mid ){
                             f = i;
                             break;
                         }
                     }
                     //printf("f = %d\n",f);
                     Point tmp;
                     tmp.y = mid;
                     tmp.x = b[f].x-( (b[f].x-b[f-].x)*(mid-b[f].y)/(b[f-].y-b[f].y) );
                     res[ cnt++] = tmp;
                     for( int i=f;i<n2;i++ ){
                         if( b[i].y<mid ){
                             res[ cnt++ ] = b[i];
                             f = i;
                         }
                         else break;
                     }
                     tmp.y = mid;
                     tmp.x = (b[ f+ ].x-b[ f ].x)*(mid-b[f].y)/(b[f+].y-b[f].y) + b[f].x;
                     res[ cnt++ ] = tmp;
                     //printf("cnt = %d\n",cnt);
                     sumArea += area( res,cnt );
                 }
                 //printf("sumarea = %lf \n",sumArea);
             /********solve a *****/
                 if( mid>aYmin ){
                     int cnt = ;
                     double newY = aYmin;
                     int f = -;
                     for( int i=;i<n1;i++ ){
                         if( a[i].y<=newY ){
                             res[ cnt++] = a[ i ];
                             f = i;
                         }
                         else break;
                     }
                     if( f==- ){}
                     else{
                         Point tmp;
                         tmp.y = newY;
                         tmp.x = (a[ f+ ].x-a[ f ].x)*(newY-a[f].y)/(a[f+].y-a[f].y) + a[f].x;
                         res[ cnt++ ] = tmp;
                     }
                     sumArea += area( res,cnt );
                 }
                 else if( mid<=amid.y ){}
                 else{
                     int cnt = ;
                     int f = -;
                     for( int i=;i<n1;i++ ){
                         if( a[i].y<=mid ){
                             f = i;
                             break;
                         }
                     }
                     Point tmp;
                     tmp.y = mid;
                     tmp.x = a[f].x-( (a[f].x-a[f-].x)*(mid-a[f].y)/(a[f-].y-a[f].y) );
                     res[ cnt++] = tmp;
                     for( int i=f;i<n1;i++ ){
                         if( a[i].y<mid ){
                             res[ cnt++ ] = a[i];
                             f = i;
                         }
                         else break;
                     }
                     tmp.y = mid;
                     tmp.x = (a[ f+ ].x-a[ f ].x)*(mid-a[f].y)/(a[f+].y-a[f].y) + a[f].x;
                     res[ cnt++ ] = tmp;
                     sumArea += area( res,cnt );
                 }
                 //printf("sumarea2 = %lf\n\n\n",sumArea);
                 if( fabs(sumArea-aim)<=eps ){
                     ansY = mid;
                     break;
                 }
                 else if( sumArea>aim ){
                     R = mid-eps;
                 }
                 else {
                     L = mid+eps;
                     ansY = mid;
                 }
             }
         }//ab可能都同时都有水
         else{
             //printf("special = %d\n",special);
             double sumArea = ;
             if( special== ){//‘1’表示只有a会有水
                 double L = amid.y;
                 double R = aYmin;
                 while( L<R ){
                     double mid = (L+R)/2.0;
                     //printf("mid = %lf\n",mid);
                     int cnt = ;
                     int f = -;
                     for( int i=;i<n1;i++ ){
                         if( a[i].y<=mid ){
                             f = i;
                             break;
                         }
                     }
                     Point tmp;
                     tmp.y = mid;
                     tmp.x = a[f].x-( (a[f].x-a[f-].x)*(mid-a[f].y)/(a[f-].y-a[f].y) );
                     res[ cnt++] = tmp;
                     for( int i=f;i<n1;i++ ){
                         if( a[i].y<mid ){
                             res[ cnt++ ] = a[i];
                             f = i;
                         }
                         else break;
                     }
                     tmp.y = mid;
                     tmp.x = (a[ f+ ].x-a[ f ].x)*(mid-a[f].y)/(a[f+].y-a[f].y) + a[f].x;
                     res[ cnt++ ] = tmp;
                     sumArea += area( res,cnt );
                     //printf("cnt = %d\n",cnt);
                     //printf("sumarea = %lf\n",sumArea);
                     if( fabs(sumArea-aim)<=eps ){
                         ansY = mid;
                         break;
                     }
                     else if( sumArea>aim ) {
                         R = mid-eps;
                     }
                     else {
                         L = mid + eps;
                         ansY = L;
                     }
                 }
             }
             else{//'2'表示只有b会有水
                 double L = bmid.y;
                 double R = bYmin;
                 //printf("L = %lf,R = %lf\n",L,R);
                 while( L<R ){
                     double mid = (L+R)/2.0;
                     //printf("mid = %lf\n",mid);
                     int cnt = ;
                     int f = -;
                     for( int i=;i<n2;i++ ){
                         if( b[i].y<=mid ){
                             f = i;
                             break;
                         }
                     }
                     Point tmp;
                     tmp.y = mid;
                     tmp.x = b[f].x-( (b[f].x-b[f-].x)*(mid-b[f].y)/(b[f-].y-b[f].y) );
                     res[ cnt++] = tmp;
                     for( int i=f;i<n2;i++ ){
                         if( b[i].y<mid ){
                             res[ cnt++ ] = b[i];
                             f = i;
                         //printf("add : i = %d\n",i);
                         }
                         else break;
                     }
                     tmp.y = mid;
                     tmp.x = (b[ f+ ].x-b[ f ].x)*(mid-b[f].y)/(b[f+].y-b[f].y) + b[f].x;
                     res[ cnt++ ] = tmp;
                     //printf("cnt = %d\n",cnt);
                     sumArea += area( res,cnt );
                     if( fabs(sumArea-aim)<=eps ){
                         ansY = mid;
                         break;
                     }
                     else if( sumArea>aim ) {
                         R = mid-eps;
                     }
                     else {
                         L = mid + eps;
                         ansY = L;
                     }
                 }
             }
         }
         printf("%.3lf\n",ansY);
     }
     return ;
 }