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-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;
struct point
{
    double x,y,r;
    point(double x=0,double y=0,double r =0):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 0;
    return x<0?-1:1;
}
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,0,sizeof(f));
        for(i = 1; i <=  n; i++)
        scanf("%lf",&p[i].r);
        p[1] = point(p[1].r,0,p[1].r);
        p[0].r = p[0].x = 0;
        double rig;
        for(i = 2; i <= n; i++)
        {
            double maxz = 0;
            int x = i-1;
            for(j = i-1 ; j >= 1 ; 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)>=0)
                {
                    maxz = max(maxz,d);
                    x = j;
                }
            }
            if(dcmp(maxz-p[i].r)<=0)
            {
                p[i].x = p[i].r;
                for(j = 1 ; j < i; j++)
                f[j] = 1;
                continue;
            }
            p[i].x = maxz;
            //cout<<i<<" "<<x<<" "<<maxz<<endl;
            for(j = x+1 ; j < i; j++)
            f[j] = 1;
        }
        int x = n;
        for(i = 1; i <= n ;i ++)
        {
            if(dcmp(p[i].r+p[i].x-rig)>0)
            {
                rig = max(rig,p[i].r+p[i].x);
                x = i;
            }
        }
        for(i = x+1 ; i <= n ;i++)
        f[i] = 1;
        int g = 0;
        for(i = 1; i <= n ;i++)
        if(f[i])
        o[++g] = i;
        cout<<g<<endl;
        for(i = 1; i <= g ;i++)
        cout<<o[i]<<endl;
    }
    return 0;
}