标签：without name sam XML sum distance ini sign extern
InputThe first line contains an integer tt indicating the total number of test cases. The following lines describe a test case.
The first line of each case contains one positive integer nn, the size of the closed path. Next nn lines, each line consists of two integers (xi,yi)(xi,yi) indicate the coordinate of the ii-th endpoint.
1≤t≤1001≤t≤100
3≤n≤1043≤n≤104
|xi|,|yi|≤104|xi|,|yi|≤104
Distances between any two adjacent endpoints are positive integers.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 nn lines, the ii-th of them should contain a number riri, rounded to 2 digits after the decimal point.
If there are several possible ways to produce the minimum sum of base area, you may output any of them.Sample Input
3 4 0 0 11 0 27 12 5 12 5 0 0 7 0 7 3 3 6 0 6 5 0 0 1 0 6 12 3 16 0 12
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\n");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\n");return; 61 } 62 } 63 } 64 printf("%.2f\n",an*acos(-1.0)); 65 for(int i=0;i<n;i++) 66 printf("%.2f\n",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\n");return; 83 } 84 double l,r; 85 findb(l,r); 86 if(l>r+eps) 87 { 88 printf("IMPOSSIBLE\n");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\n",get(l)*acos(-1.0)); 100 for(int i=0;i<n;i++) 101 { 102 printf("%.2f\n",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 }
标签：without name sam XML sum distance ini sign extern
原文地址：http://www.cnblogs.com/quintessence/p/6940822.html
