bzoj1857: [Scoi2010]传送带--三分套三分

三分套三分模板

貌似只要是单峰函数就可以用三分求解

 #include<stdio.h>
 #include<string.h>
 #include<algorithm>
 #include<math.h>
 #define eps 1e-9
 using namespace std;
 struct node{
     double x,y;
 }a,b,c,d;
 double p,q,r;

 inline node get(node a, node b, double p){
     node ans;
     ans.x=a.x+(b.x-a.x)*p;
     ans.y=a.y+(b.y-a.y)*p;
     return ans;
 }

 inline double dist(node x, node y){
     return sqrt((x.x-y.x)*(x.x-y.x)+(x.y-y.y)*(x.y-y.y));
 }

 inline double calc(node x, node y){
     return dist(a,x)/p+dist(x,y)/r+dist(y,d)/q;
 }

 inline double solve(node t){
     double l=0.0,r=1.0,ans;
     while (fabs(l-r)>eps){
         , m2=m1+(r-l)*;
         node x=get(c,d,m1), y=get(c,d,m2);
         if (calc(t,x)<calc(t,y)) ans=m1,r=m2; else ans=m2, l=m1;
     }
     node x=get(c,d,ans);
     return calc(t,x);
 }

 int main(){
     scanf("%lf%lf%lf%lf", &a.x, &a.y, &b.x, &b.y);
     scanf("%lf%lf%lf%lf", &c.x, &c.y, &d.x, &d.y);
     scanf("%lf%lf%lf", &p, &q, &r);
     double l=0.0,r=1.0,ans;
     while (fabs(l-r)>eps){
         , m2=m1+(r-l)*;
         node x=get(a,b,m1), y=get(a,b,m2);
         if (solve(x)<solve(y)) ans=m1,r=m2; else ans=m2,l=m1;
     }
     node x=get(a,b,ans);
     printf("%.2lf\n", solve(x));
     ;
 }