HDOJ 4355 —— 三分

Party All the Time

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2383    Accepted Submission(s): 800

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.

 

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 : X_i,W_i, representing the location and the weight of the i-th spirit. ( |x_i|<=10⁶, 0<w_i<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

140.6 53.9 105.1 78.4 10

 

Sample Output

Case #1: 832

 

Author

Enterpaise@UESTC_Goldfinger

 

Source

2012 Multi-University Training Contest 6

 

Recommend

zhuyuanchen520

 

直接三分即可，G++ 765MS ， C++会TLE.。。。

#include <cstdio>#include <cmath>#include <algorithm>#include <iostream>#include <cstring>#include <map>#include <string>#include <stack>#include <cctype>#include <vector>#include <queue>#include <set>#include <utility>using namespace std;//#define Online_Judge#define outstars cout << "***********************" << endl;#define clr(a,b) memset(a,b,sizeof(a))#define lson l , mid  , rt << 1#define rson mid + 1 , r , rt << 1 | 1//#define mid ((l + r) >> 1)#define mk make_pair#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)const int MAXN = 50000 + 500;const long long LLMAX = 0x7fffffffffffffffLL;const long long LLMIN = 0x8000000000000000LL;const int INF = 0x3f3f3f3f;const int IMIN = 0x80000000;const double e = 2.718281828;#define eps 1e-8#define DEBUG 1#define mod 1000000007typedef long long LL;const double PI = acos(-1.0);typedef double D;typedef pair<int , int> pi;///#pragma comment(linker, "/STACK:102400000,102400000")__int64 a[10050];int n ;D x[MAXN] , w[MAXN];D work(D xx){    D ans = 0.0;    FORR(i ,  0 , n)    {        D t = abs(xx - x[i]);        ans += t * t * t * w[i];    }    return ans;}LL triSearch(D l , D r){    D mid , midmid , valueofmid , valueofmidmid;    while(r - l > eps)    {        mid = (l + r) / 2;        midmid = (mid + r) / 2;        valueofmid = work(mid);        valueofmidmid = work(midmid);        if(valueofmid <= valueofmidmid)r = midmid;        else l = mid;    }    return LL(valueofmid + 0.5);}int main(){    int  t;    cin >> t;    FORR(kase , 1 , t)    {        scanf("%d" ,&n);        FOR(i ,0 , n)        {            scanf("%lf%lf" , &x[i] , &w[i]);        }        printf("Case #%d: " , kase);        if(n == 1)printf("%I64d\n" , (LL)(work(x[0]) + 0.5));        else printf("%I64d\n" , (LL)(triSearch(x[0] , x[n -1]) + 0.5));    }    return 0;}