2012 Multi-University Training Contest 6-1006 hdu4355 Party All the Time
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这个题就是求一个点p,使得(xp - xi ) ^ 3 (i = 1,2,..n)的和最小,通过大家总结这是一个单峰函数(具体是为什么我也不知道),通过三分法可以得到正解。
以下是一个很好的三分法的模板~~
#include <iostream>#include <cstdio>#include <cmath>#include <iomanip>using namespace std;typedef double D;const D eps = 1e-8;D x[50005];int n;D w[50005];D func(D y){ D temp = 0.0; for(int i = 1;i <= n;i++) { if(x[i] < y) temp += (y-x[i] )* (y-x[i] ) * (y-x[i] ) * w[i]; else if(x[i] > y) temp += (x[i]-y)* (x[i]-y) * (x[i]-y) * w[i]; } return temp;}D minn(D a,D b){ if(a < b) return a; else return b;}D Ternary_Search(D lb, D rb){ D x1,x2; while(fabs(1-lb/rb)>eps) { x1 = lb + (rb-lb)/3,x2 = rb-(rb-lb)/3; if(func(x1)<func(x2)) rb = x2; else lb = x1; } return func(lb);}int main(){ int t,cas =1; scanf("%d",&t); while(t--) { scanf("%d",&n); for(int i = 1;i <= n;i++) { scanf("%lf%lf",&x[i],&w[i]); } D ans = Ternary_Search(x[1],x[n]); printf("Case #%d: ",cas++); cout <<fixed << setprecision(0)<< ans<< endl; } return 0;}
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