Rebuild

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2483 Accepted Submission(s): 548

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 endpoint i−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 r

i 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

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

題目大意：

有 n 個點，其中這 n 個點能夠組成一個凸 n 邊形，而且是按照順序的，這樣就不用排序了，如果不給定順序的話，應該按照極角進行排序，現在以每個點為圓心，保證兩個相鄰點為圓心的圓是相切的，現在求的是所有的圓的面積和，並且將每個圓的半徑輸出。

解題思路：

因為保證相鄰圓是相切的，所以我們設第一個圓的半徑為 r1 ，那麼有如下式子成立：
r1+r2=d1
r2+r3=d2
...
rn+r1=dn

然後我們來解這個方程，解的過程中就會發現有一個小規律，跟 n 的奇偶有關係，即：當 n 為奇數的時候，就會有唯一解能夠解出 r1，r1=d1−d2+d3+...+dn2
如果是偶數的話，我們發現解不出這個方程，那麼怎麼做的，因為求的是面積，所以最終的結果一定是關於

HDU 5531 Rebuild（三分）——2015ACM/ICPC亞洲區長春站

Rebuild

HDU 5531 Rebuild（三分）——2015ACM/ICPC亞洲區長春站

hdu 5531 Rebuild（三分）

BZOJ3203 SDOI2013保護出題人（三分）

ACM_求第k大數（三分）

（三分）TOJ3777 Function Problem

#LOJ10013 曲線（三分）

poj3301 Texas Trip （三分）

poj 3737 UmBasketella （三分）

HPUOJ 題目1079 假幣問題（三分）

@2015ACM/ICPC亞洲區長春站-重現賽（感謝東北師大） @HDU5534 (Dp,完全揹包)

HDU 5532 Almost Sorted Array(最長非遞減子序列模板題)——2015ACM/ICPC亞洲區長春站

【hdu 5527】 [2015ACM/ICPC亞洲區長春站] Too Rich　貪心

[hdu 5521][2015ACM/ICPC亞洲區長春站] House Building

5120 Intersection（簡單幾何）——2014ACM/ICPC亞洲區北京站

[hdu 5534]2015ACM/ICPC亞洲區長春站　Partial Tree　　完全揹包

【hdu5538】【2015ACM/ICPC亞洲區長春站】House Building　題意＆題解＆程式碼

【套題】2015ACM/ICPC亞洲區長春站 HDU5532 5533 5534 5536 5538

【hdu5534】【2015ACM/ICPC亞洲區長春站】Partial Tree 題意＆題解＆程式碼

HDU 3400 Line belt （三分套三分）

HDU-2438-Turn the corner（三分+思維）

HDU 5531 Rebuild（三分）——2015ACM/ICPC亞洲區長春站

Rebuild

相關推薦