（三分）HDU5531 Rebuild

Archaeologists find ruins of Ancient ACM Civilization, and they want to rebuild it.

The ruins form a closed path on an x-y plane, which has

n

n

InputThe first line contains an integer $t$ OutputIf such answer doesn't exist, then print on a single line "IMPOSSIBLE" (without the quotes). Otherwise, in the first line print the minimum sum of base area, and then print $n$ Sample Input

Sample Output

988.82
3.75
7.25
12.75
9.25
157.08
6.00
1.00
2.00
3.00
0.00
IMPOSSIBLE

按点的数目奇偶分情况，奇数情况直接可以解方程，偶数情况其他所有数可以用r1表示，确定r1范围，三分即可。（这题精度有点神奇……）

  1 #include <iostream>
  2 #include <string>
  3 #include <algorithm>
  4 #include <cstring>
  5 #include <cstdio>
  6 #include <cmath>
  7 #include <queue>
  8 #include <set>
  9 #include <map>
 10 #include <list>
 11 #include <vector>
 12 #include <stack>
 13 #define mp make_pair
 14 //#define P make_pair
 15 #define MIN(a,b) (a>b?b:a)
 16 //#define MAX(a,b) (a>b?a:b)
 17 typedef long long ll;
 18 typedef unsigned long long ull;
 19 const int MAX=1e4+5;
 20 const int MAX_V=1e3+5;
 21 const ll INF=4e18+5;
 22 const double M=4e18;
 23 using namespace std;
 24 const int MOD=1e9+7;
 25 typedef pair<ll,int> pii;
 26 const double eps=1e-7;
 27 #define rank rankk
 28 int t,n;
 29 double x[MAX],y[MAX],d[MAX],r[MAX],g[MAX];
 30 int findb(double &l,double &r)
 31 {
 32     l=0,r=min(d[0],d[n-1]);
 33     for(int i=0;i<n-1;i++)
 34         if(i&1)
 35             l=max(l,g[i]);
 36         else
 37             r=min(r,g[i]);
 38 }
 39 void solve_odd()
 40 {
 41     double sum=0.0,an=0.0;
 42     for(int i=0;i<n;i++)
 43         if(i&1)
 44             sum-=d[i];
 45         else
 46             sum+=d[i];
 47     r[0]=sum/2.0;
 48     if(r[0]<-eps)
 49     {
 50         printf("IMPOSSIBLE
");return;
 51     }
 52     else
 53     {
 54         for(int i=0;i<n;i++)
 55         {
 56             an+=r[i]*r[i];
 57             r[i+1]=d[i]-r[i];
 58             if(i+1!=n&&r[i+1]<-eps)
 59             {
 60                 printf("IMPOSSIBLE
");return;
 61             }
 62         }
 63     }
 64     printf("%.2f
",an*acos(-1.0));
 65     for(int i=0;i<n;i++)
 66         printf("%.2f
",r[i]);
 67     return;
 68 }
 69 double get(double m) {
 70     double re=m*m,r=m;
 71     for (int i=0;i<n-1;i++) {
 72         r=d[i]-r;
 73         re+=r*r;
 74     }
 75     return re;
 76 }
 77 
 78 void solve_even()
 79 {
 80     if(abs(g[n-1])>eps)
 81     {
 82         printf("IMPOSSIBLE
");return;
 83     }
 84     double l,r;
 85     findb(l,r);
 86     if(l>r+eps)
 87     {
 88         printf("IMPOSSIBLE
");return;
 89     }
 90     while(r-l>eps)
 91     {
 92         double L,R;
 93         L=(2.0*l+r)/3.0;R=(l+2.0*r)/3.0;
 94         if(get(L)>=get(R)+eps)
 95             l=L;
 96         else
 97             r=R;
 98     }
 99     printf("%.2f
",get(l)*acos(-1.0));
100     for(int i=0;i<n;i++)
101     {
102         printf("%.2f
",l);
103         l=d[i]-l;
104     }
105 }
106 int main()
107 {
108     scanf("%d",&t);
109     while(t--)
110     {
111         scanf("%d",&n);
112         for(int i=0;i<n;i++)
113         {
114             scanf("%lf%lf",&x[i],&y[i]);
115             if(i)
116                 d[i-1]=sqrt((x[i]-x[i-1])*(x[i]-x[i-1])+(y[i]-y[i-1])*(y[i]-y[i-1]));
117         }
118         d[n-1]=sqrt((x[n-1]-x[0])*(x[n-1]-x[0])+(y[n-1]-y[0])*(y[n-1]-y[0]));
119         g[0]=d[0];
120         for(int i=1;i<n;i++)
121             if(i&1)
122                 g[i]=g[i-1]-d[i];
123             else
124                 g[i]=g[i-1]+d[i];
125         if(n&1)
126             solve_odd();
127         else
128             solve_even();
129     }
130     return 0;
131 }

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原文地址：https://www.cnblogs.com/quintessence/p/6940822.html