Rebuild

https://blog.csdn.net/LuRiCheng/article/details/70135430

自己懶得寫，轉載的別人的解釋挺好很詳細

/**
 * ============================
 * @title:    changcun 2015 PE
 * @Author:   kuroko
 * @Version:  1.0 
 * @DateTime: 2018-08-31 09:31:59
 * ============================
 */
#include<bits/stdc++.h>
using namespace
 
 std;
const int maxn = 1e4+10;
const double pi =acos(-1.0);
const double eps=1e-6;
int n,x[maxn],y[maxn];
double d[maxn],r[maxn];
int sgn(double t)
{
    return (t>eps)-(t<-eps);
}
double f(double a, double b, double c, double t)
{
    return a*t*t + b*t + c;
}
void input()
{
    scanf(
 
"%d",&n);
    for(int i=0;i<n;i++)
        scanf("%d%d",&x[i],&y[i]);
}
void solve()
{
    x[n]=x[0];
    y[n]=y[0];
    for(int i=0;i<n;i++)
        d[i]=sqrt((x[i+1]-x[i])*(x[i+1]-x[i])+(y[i+1]-y[i])*(y[i+1]-y[i]));
    if(n&1)//奇數
    {
        double all=0,other=0,ans=0;
        
 
for(int i=0;i<n;i++)
        {
            all+=d[i];
            if(i&1)
              other+=d[i];
        }
        r[0]=all/2-other;
        if(sgn(r[0])<0)
        {
            printf("IMPOSSIBLE\n");
            return;
        }
        ans+=r[0]*r[0];
        for(int i=1;i<n;i++)
        {
            r[i]=d[i-1]-r[i-1];
            if(sgn(r[i])<0)
            {
                printf("IMPOSSIBLE\n");
                return;
            }
            ans+=r[i]*r[i];
        }
        printf("%.2lf\n",pi*ans);
        for(int i=0;i<n;i++)
            printf("%.2lf\n",r[i]);
    }
    else
    {
        double temp=0;
        for(int i=0;i<n;i+=2)
            temp+=d[i]-d[i+1];
        if(sgn(temp)!=0)
        {
            printf("IMPOSSIBLE\n");
            return ;
        }
        int k=1;
        double b=0,L=0,R=d[0];
        double A=1,B=0,C=0;
        //計算的是ri=b+k*r0(k=1或-1)
        for(int i=1;i<n;i++)
        {
            k=-k;
            b=-b+d[i-1];
            if(k==1)
                L=max(L,-b);
            else//減去的時候
                R=min(R,b);
            A++;
            B+=2*k*b;
            C+=b*b;
        }
        if(sgn(L-R)>0)
        {
            printf("IMPOSSIBLE\n");
            return;
        }
        double mid=-B/(2*A),ans;
        if(sgn(L-mid)>=0)
        {
            ans=f(A,B,C,L);
            r[0]=L;
        }
        if(sgn(R-mid)<=0)
        {
            ans=f(A,B,C,R);
            r[0]=R;
        }
        if(sgn(mid-L)>=0&&sgn(mid-R)<=0)
        {
            ans=f(A,B,C,mid);
            r[0]=mid;
        }
        printf("%.2f\n",pi*ans); 
        for(int i=0;i<n;i++)
        {
            if(i)
                r[i]=d[i-1]-r[i-1];
            printf("%.2lf\n",r[i]);
        }
    }
}
int main(int argc, char const *argv[])
{
    freopen("in.txt","r",stdin);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        input();
        solve();
    }
    return 0;
}

ACM Changchun 2015-E Rebuild