三分法:Party all the time
3103 ワード
Party All the Time
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 829 Accepted Submission(s): 308
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 S
3*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.
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<=ii,W
i, representing the location and the weight of the i-th spirit. ( |x
i|<=10
6, 0i<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
Sample Output
凸関数、直接三分法で解く
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 829 Accepted Submission(s): 308
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 S
3*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.
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
i, representing the location and the weight of the i-th spirit. ( |x
i|<=10
6, 0
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
凸関数、直接三分法で解く
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define EPS 1e-7
#define MAXN 50050
double x[MAXN], w[MAXN];
int n;
double Cal(double pos){
double ans = 0;
for(int i = 1; i <= n; i ++){
double tmp = abs(pos - x[i]);
ans += w[i] * tmp * tmp * tmp;
}
return ans;
}
double Solve(){
double Left, Right;
double mid, midmid;
double mid_value, midmid_value;
Left = x[1];
Right = x[n];
while(Left + EPS < Right){
mid = (Left + Right) / 2;
midmid = (mid + Right) / 2;
mid_value = Cal(mid);
midmid_value = Cal(midmid);
if(mid_value <= midmid_value)
Right = midmid;
else
Left = mid;
}
return Cal(Left);
}
int main(){
int t;
int cnt = 0;
scanf("%d", &t);
while(t --){
cnt ++;
scanf("%d", &n);
for(int i = 1; i <= n; i ++){
scanf("%lf %lf", &x[i], &w[i]);
}
printf("Case #%d: %.0lf
", cnt, Solve());
}
return 0;
}