POJ_1064_Cable_master_(二分,假定一个解并判断是否可行)
描述
http://poj.org/problem?id=1064
有n条绳子,长度分别为l[i].如果从它们中切割出k条长度相同的绳子的话,这k条绳子每条最长能有多少?
Cable master
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 35271 | Accepted: 7515 |
Description
Inhabitants of the Wonderland have decided to hold a regional programming contest. The Judging Committee has volunteered and has promised to organize the most honest contest ever. It was decided to connect computers for the contestants using a "star" topology - i.e. connect them all to a single central hub. To organize a truly honest contest, the Head of the Judging Committee has decreed to place all contestants evenly around the hub on an equal distance from it.
To buy network cables, the Judging Committee has contacted a local network solutions provider with a request to sell for them a specified number of cables with equal lengths. The Judging Committee wants the cables to be as long as possible to sit contestants as far from each other as possible.
The Cable Master of the company was assigned to the task. He knows the length of each cable in the stock up to a centimeter,and he can cut them with a centimeter precision being told the length of the pieces he must cut. However, this time, the length is not known and the Cable Master is completely puzzled.
You are to help the Cable Master, by writing a program that will determine the maximal possible length of a cable piece that can be cut from the cables in the stock, to get the specified number of pieces.
To buy network cables, the Judging Committee has contacted a local network solutions provider with a request to sell for them a specified number of cables with equal lengths. The Judging Committee wants the cables to be as long as possible to sit contestants as far from each other as possible.
The Cable Master of the company was assigned to the task. He knows the length of each cable in the stock up to a centimeter,and he can cut them with a centimeter precision being told the length of the pieces he must cut. However, this time, the length is not known and the Cable Master is completely puzzled.
You are to help the Cable Master, by writing a program that will determine the maximal possible length of a cable piece that can be cut from the cables in the stock, to get the specified number of pieces.
Input
The
first line of the input file contains two integer numb ers N and K,
separated by a space. N (1 = N = 10000) is the number of cables in the
stock, and K (1 = K = 10000) is the number of requested pieces. The
first line is followed by N lines with one number per line, that specify
the length of each cable in the stock in meters. All cables are at
least 1 meter and at most 100 kilometers in length. All lengths in the
input file are written with a centimeter precision, with exactly two
digits after a decimal point.
Output
Write
to the output file the maximal length (in meters) of the pieces that
Cable Master may cut from the cables in the stock to get the requested
number of pieces. The number must be written with a centimeter
precision, with exactly two digits after a decimal point.
If it is not possible to cut the requested number of pieces each one being at least one centimeter long, then the output file must contain the single number "0.00" (without quotes).
If it is not possible to cut the requested number of pieces each one being at least one centimeter long, then the output file must contain the single number "0.00" (without quotes).
Sample Input
4 11 8.02 7.43 4.57 5.39
Sample Output
2.00
Source
分析
二分.
条件C(x)=可以得到k条长度为x的绳子,求满足C(x)的x的最小值.求解这样的最大化或最小化问题:"假定一个解并判断是否可行".(普通二分查找值的条件C(x)=v).
注意:
1.关于精度,这个貌似不太好估算,所以就循环100次咯.
1 #include<cstdio> 2 #include<algorithm> 3 using std :: max; 4 5 const int maxn=10005; 6 int n,k; 7 double maxl,l[maxn]; 8 9 bool judge(double x) 10 { 11 int sum=0; 12 for(int i=1;i<=n;i++) 13 { 14 sum+=(int)(l[i]/x); 15 } 16 return sum>=k; 17 } 18 19 double bsearch(double x,double y) 20 { 21 for(int i=0;i<300;i++) 22 { 23 double m=x+(y-x)/2; 24 if(judge(m)) x=m; 25 else y=m; 26 } 27 return x; 28 } 29 30 void init() 31 { 32 scanf("%d%d",&n,&k); 33 for(int i=1;i<=n;i++) 34 { 35 scanf("%lf",&l[i]); 36 maxl=max(maxl,l[i]); 37 } 38 } 39 40 void solve() 41 { 42 printf("%.2f\n",floor(bsearch(0,maxl)*100)/100); 43 } 44 45 int main() 46 { 47 freopen("Cable.in","r",stdin); 48 freopen("Cable.out","w",stdout); 49 init(); 50 solve(); 51 fclose(stdin); 52 fclose(stdout); 53 return 0; 54 }