Party All the Time
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 926 Accepted Submission(s): 341
Problem Description
In
the Dark forest, there is a Fairy kingdom where all the spirits will go
together and Celebrate the harvest every year. But there is one thing
you may not know that they hate walking so much that they would prefer
to stay at home if they need to walk a long way.According to our
observation,a spirit weighing W will increase its unhappyness for S3*W units if it walks a distance of S kilometers.
Now give you every spirit's weight and location,find the best place to celebrate the harvest which make the sum of unhappyness of every spirit the least.
Now give you every spirit's weight and location,find the best place to celebrate the harvest which make the sum of unhappyness of every spirit the least.
Input
The
first line of the input is the number T(T<=20), which is the number
of cases followed. The first line of each case consists of one integer
N(1<=N<=50000), indicating the number of spirits. Then comes N
lines in the order that x[i]<=x[i+1] for all i(1<=i<N). The i-th line contains two real number : Xi,Wi, representing the location and the weight of the i-th spirit. ( |xi|<=106, 0<wi<15 )
Output
For
each test case, please output a line which is "Case #X: Y", X means the
number of the test case and Y means the minimum sum of unhappyness
which is rounded to the nearest integer.
Sample Input
1
4
0.6 5
3.9 10
5.1 7
8.4 10
Sample Output
Case #1: 832
Author
Enterpaise@UESTC_Goldfinger
Source
因为 s^3 *w s加了 绝对值 所以 是一个 凸性 函数 可以用 三分做
三分搜索
1 #include<stdio.h> 2 #include<iostream> 3 #include<algorithm> 4 #include<cstring> 5 #include<cmath> 6 #include<queue> 7 #include<set> 8 #include<map> 9 #define Min(a,b) a>b?b:a 10 #define Max(a,b) a>b?a:b 11 #define CL(a,num) memset(a,num,sizeof(a)); 12 #define inf 9999999 13 #define maxn 50010 14 #define eps 1e-5 15 #define ll long long 16 using namespace std; 17 double a[maxn],w[maxn]; 18 int n; 19 double f(double pos) 20 { 21 int i; 22 double ans = 0; 23 for( i = 0 ; i < n ; ++i) 24 { 25 double k = fabs(a[i] - pos); 26 27 ans += k*k*k*w[i]; 28 } 29 return ans ; 30 } 31 double solve(double l,double r) 32 { 33 34 double mid,midmid; 35 while(r - l > eps) 36 { 37 mid = (l + r)/2.0; 38 midmid = (r + mid)/2.0; 39 40 41 if(f(mid) <= f(midmid))// f 就算函数值 42 r = midmid; 43 else l = mid; 44 } 45 return f(l); 46 } 47 int main() 48 { 49 int t,i; 50 scanf("%d",&t); 51 int cas = 0; 52 while(t--) 53 { 54 scanf("%d",&n); 55 for(i = 0 ;i < n;++i ) 56 { 57 scanf("%lf %lf",&a[i],&w[i]); 58 } 59 double ans = solve(a[0],a[n - 1]); 60 printf("Case #%d: %.0lf\n",++cas,ans); 61 } 62 }