標籤:codeforces 刷題
類比退火果然是一個很高端的東西,思路神馬的完全搞不懂啊~
題目大意:
給出三個圓,求一點到這三個圓的兩切線的夾角相等。
解題思路:
對於這個題來說還是有多種思路的 ,不過都搞不明白~~ /害羞臉
用類比退火來解也是一件賭人品的事,因為退火的過程設計的不合理,WA妥妥的。
其實我也是學了一點點,還不是太明白啊~~
下面是代碼:
#include <set>#include <map>#include <queue>#include <math.h>#include <vector>#include <string>#include <stdio.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <cctype>#include <algorithm>#define eps 1e-6#define pi acos(-1.0)#define inf 107374182#define inf64 1152921504606846976#define lc l,m,tr<<1#define rc m + 1,r,tr<<1|1#define zero(a) fabs(a)<eps#define iabs(x) ((x) > 0 ? (x) : -(x))#define clear1(A, X, SIZE) memset(A, X, sizeof(A[0]) * (min(SIZE,sizeof(A))))#define clearall(A, X) memset(A, X, sizeof(A))#define memcopy1(A , X, SIZE) memcpy(A , X ,sizeof(X[0])*(SIZE))#define memcopyall(A, X) memcpy(A , X ,sizeof(X))#define max( x, y ) ( ((x) > (y)) ? (x) : (y) )#define min( x, y ) ( ((x) < (y)) ? (x) : (y) )using namespace std;int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};struct circle{ double x,y,r;}cir[3];double dis(double x,double y,double xx,double yy){ return sqrt((x-xx)*(x-xx)+(y-yy)*(y-yy));}double f(double x,double y){ double tmp[3]; for(int i=0;i<3;i++) tmp[i]=dis(x,y,cir[i].x,cir[i].y)/cir[i].r; //視角一半的sin值 double ans=0; for(int i=0;i<3;i++) ans+=(tmp[i]-tmp[(i+1)%3])*(tmp[i]-tmp[(i+1)%3]); return ans;}int main(){ double x=0,y=0; for(int i=0;i<3;i++) { scanf("%lf%lf%lf",&cir[i].x,&cir[i].y,&cir[i].r); x+=cir[i].x/3; y+=cir[i].y/3; } double step=2; while(step>eps) { double tmp=f(x,y); int tag=-1; for(int i=0;i<4;i++) { double cnt=f(x+dir[i][0]*step,y+dir[i][1]*step); if(cnt<tmp) { tmp=cnt; tag=i; } } if(tag==-1) step/=2; else { x=x+dir[tag][0]*step; y=y+dir[tag][1]*step; } } if(f(x,y)<eps) printf("%.5lf %.5lf\n",x,y); return 0;}
Codeforces Beta Round #2 C. Commentator problem