poj1819Disks

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题意：从左到右按顺序给你n个圆的半径，把左右两边想象成两堵墙的话，就是左右两边向里挤压，问哪些圆是对最后的宽度不影响。

刚开始理解错了，。。以为怎么放圆使宽度最小。。

这样就可以尽量使每个圆向左靠，找出当前圆与最左相切的圆，他们之间的那些圆肯定就是可以消除的，特判一下最左和最右就可以了。

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 1010
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 struct point
 {
     double x,y,r;
     point(double x=,double y=,double r =):x(x),y(y),r(r){}
 }p[N];
 int o[N];
 bool f[N];
 typedef point pointt;
 pointt operator -(point a,point b)
 {
     return point(a.x-b.x,a.y-b.y);
 }
 int dcmp(double x)
 {
     if(fabs(x)<eps) return ;
     return x<?-:;
 }
 double dis(point a)
 {
     return sqrt(a.x*a.x+a.y*a.y);
 }
 int main()
 {
     int n,i,j;
     while(scanf("%d",&n)!=EOF)
     {
         memset(f,,sizeof(f));
         for(i = ; i <=  n; i++)
         scanf("%lf",&p[i].r);
         p[] = point(p[].r,,p[].r);
         p[].r = p[].x = ;
         double rig;
         for(i = ; i <= n; i++)
         {
             double maxz = ;
             int x = i-;
             for(j = i- ; j >=  ; j--)
             {
                 double d = sqrt((p[j].r+p[i].r)*(p[j].r+p[i].r)-(p[j].r-p[i].r)*(p[j].r-p[i].r))+p[j].x;
                 if(dcmp(d-maxz)>=)
                 {
                     maxz = max(maxz,d);
                     x = j;
                 }
             }
             if(dcmp(maxz-p[i].r)<=)
             {
                 p[i].x = p[i].r;
                 for(j =  ; j < i; j++)
                 f[j] = ;
                 continue;
             }
             p[i].x = maxz;
             //cout<<i<<" "<<x<<" "<<maxz<<endl;
             for(j = x+ ; j < i; j++)
             f[j] = ;
         }
         int x = n;
         for(i = ; i <= n ;i ++)
         {
             if(dcmp(p[i].r+p[i].x-rig)>)
             {
                 rig = max(rig,p[i].r+p[i].x);
                 x = i;
             }
         }
         for(i = x+ ; i <= n ;i++)
         f[i] = ;
         int g = ;
         for(i = ; i <= n ;i++)
         if(f[i])
         o[++g] = i;
         cout<<g<<endl;
         for(i = ; i <= g ;i++)
         cout<<o[i]<<endl;
     }
     return ;
 }