Rebuild
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 521 Accepted Submission(s): 125
Problem Description
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 endpoints. The endpoints locate on (x1,y1), (x2,y2), …,(xn,yn) respectively. Endpoint i and endpointi−1 are adjacent for 1<i≤n, also endpoint 1 and endpoint n are adjacent. Distances between any two adjacent endpoints are positive integers.
To rebuild, they need to build one cylindrical pillar at each endpoint, the radius of the pillar of endpoint i is ri. All the pillars perpendicular to the x-y plane, and the corresponding endpoint is on the centerline of it. We call two pillars are adjacent if and only if two corresponding endpoints are adjacent. For any two adjacent pillars, one must be tangent externally to another, otherwise it will violate the aesthetics of Ancient ACM Civilization. If two pillars are not adjacent, then there are no constraints, even if they overlap each other.
Note that ri must not be less than 0 since we cannot build a pillar with negative radius and pillars with zero radius are acceptable since those kind of pillars still exist in their neighbors.
You are given the coordinates of n endpoints. Your task is to find r1,r2,…,rn which makes sum of base area of all pillars as minimum as possible.
For example, if the endpoints are at (0,0), (11,0), (27,12), (5,12), we can choose (r1, r2, r3, r4)=(3.75, 7.25, 12.75, 9.25). The sum of base area equals to 3.752π+7.252π+12.752π+9.252π=988.816…. Note that we count the area of the overlapping parts multiple times.
If there are several possible to produce the minimum sum of base area, you may output any of them.
Input
The first line contains an integer t indicating the total number of test cases. The following lines describe a test case.
The first line of each case contains one positive integer n, the size of the closed path. Next n lines, each line consists of two integers (xi,yi) indicate the coordinate of the i-th endpoint.
1≤t≤100
3≤n≤104
|xi|,|yi|≤104
Distances between any two adjacent endpoints are positive integers.
Output
If 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 lines, the i-th of them should contain a number ri, 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
Source
2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）
【题意】：
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cmath>
5 #include<algorithm>
6 #define eps 1e-8
7 #define PI acos(-1.0)
8 #define LL long long
9 #define maxn 100100
10 #define IN freopen("in.txt","r",stdin);
11 using namespace std;
12
13
14 struct Point {
15 double x,y;
16 };
17
18 int sign(double x) {
19 if(fabs(x)<eps) return 0;
20 return x<0? -1:1;
21 }
22
23 double Distance(Point p1,Point p2) {
24 return sqrt((p1.x-p2.x)*(p1.x-p2.x) + (p1.y-p2.y)*(p1.y-p2.y));
25 }
26
27
28 int n;
29 Point p[maxn];
30 double d[maxn];
31
32 int check(double r)
33 {
34 if(sign(r) < 0) return 0;
35
36 for(int i=2 ;i<=n; i++) {
37 r = d[i]-r;
38 if(sign(r) < 0) return 0;
39 }
40 return 1;
41 }
42
43
44 /*三分法求面积最小时的r*/
45 double a,b,c;
46 double cal(double x) {
47 return a*x*x + b*x + c;
48 }
49
50 double solve(double low, double high)
51 {
52 while(high-low > eps){
53 double mid = (high+low)/2.0;
54 double midmid = (mid+high)/2.0;
55 if(sign(cal(midmid)-cal(mid)) > 0) high = midmid;
56 else low = mid;
57 }
58
59 return low;
60 }
61
62
63 int main(int argc, char const *argv[])
64 {
65 //IN;
66
67 int t;scanf("%d",&t);
68 while(t--)
69 {
70 double r = 0;
71 memset(p, 0, sizeof(p));
72 memset(d, 0, sizeof(d));
73 int flag = 1;
74
75 scanf("%d", &n);
76 for(int i=1 ;i<=n; i++)
77 scanf("%lf %lf", &p[i].x, &p[i].y);
78 d[1]=Distance(p[1], p[n]);
79 for(int i=2; i<=n; i++)
80 d[i] = Distance(p[i], p[i-1]);
81
82 if(n%2 != 0){
83 r = d[1];
84 for(int i=2; i<=n; i++){
85 if(i%2 == 0) r += d[i];
86 else r -= d[i];
87 }
88 r /= 2.0;
89 flag = check(r);
90 }
91
92 else{
93 double tmp = 0;
94 for(int i=1; i<=n; i++){
95 if(i%2 != 0) tmp += d[i];
96 else tmp -= d[i];
97 }
98 if(sign(tmp) != 0) {flag=0; goto end;}
99
100 /*半径平方和的二次表达式*/
101 a=1.0; b=0; c=0;
102 tmp = 0;
103 for(int i=2; i<=n; i++){
104 a += 1.0;
105 tmp = d[i]-tmp;
106 c += tmp*tmp;
107 if(i%2 == 0) b -= 2.0*tmp;
108 else b += 2.0*tmp;
109 }
110
111 /*
112 错误地以为最低点为唯一取值点,若最低点不符合则无解.
113 r=(-b)/(2.0*a);
114 if(sign(r)<0) r=0;
115 flag = check(r);
116 */
117
118 /*检查每条边计算三分时r的可行范围*/
119 double low=0, high = d[1];
120 tmp = 0;
121 for(int i=2; i<=n; i++){
122 if(i%2 == 0){
123 tmp += d[i];
124 high = min(high, tmp);
125 }
126 else{
127 tmp -= d[i];
128 low = max(low, tmp);
129 }
130 }
131
132 if(low > high) flag = 0;
133 else{
134 r = solve(low, high);
135 flag = check(r);
136 }
137 }
138
139 end:
140 if(!flag) printf("IMPOSSIBLE\n");
141 else{
142 double R[maxn];
143 R[1]=r;
144 double temp = 0;
145 for(int i=2; i<=n; i++){
146 temp = d[i]-R[i-1];
147 R[i] = temp;
148 }
149 double ans = 0;
150 for(int i=1; i<=n; i++){
151 //ans += R[i]*R[i]*PI;精度问题
152 ans += R[i]*R[i];
153 }
154
155 printf("%.2lf\n", ans*PI);
156 for(int i=1; i<=n; i++){
157 printf("%.2lf\n", R[i]);
158 }
159 }
160
161 }
162
163 return 0;
164 }