2020 cattle off winter algorithm base Camp 5-B ground beef clan match

 

 

Thinking

　　Third of the template title, did not look too long because the simulated annealing parameter adjustment will not only half the water points had come to play a positive solution. .

 

 CODE

 

  1 #include <bits/stdc++.h>
  2 #define dbg(x) cout << #x << "=" << x << endl
  3 #define eps 1e-8
  4 
  5 using namespace std;
  6 typedef long long LL;
  7 
  8 template<class T>inline void read(T &res)
  9 {
 10     char c;T flag=1;
 11     while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0';
 12     while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag;
 13 }
 14 
 15 namespace _buff {
 16     const size_t BUFF = 1 << 19;
 17     char ibuf[BUFF], *ib = ibuf, *ie = ibuf;
 18     char getc() {
 19         if (ib == ie) {
 20             ib = ibuf;
 21             ie = ibuf + fread(ibuf, 1, BUFF, stdin);
 22         }
 23         return ib == ie ? -1 : *ib++;
 24     }
 25 }
 26 
 27 int qread() {
 28     using namespace _buff;
 29     int ret = 0;
 30     bool pos = true;
 31     char c = getc();
 32     for (; (c < '0' || c > '9') && c != '-'; c = getc()) {
 33         assert(~c);
 34     }
 35     if (c == '-') {
 36         pos = false;
 37         c = getc();
 38     }
 39     for (; c >= '0' && c <= '9'; c = getc()) {
 40         ret = (ret << 3) + (ret << 1) + (c ^ 48);
 41     }
 42     return pos ? ret : -ret;
 43 }
 44 
 45 const int maxn = 1e5 + 7;
 46 
 47 int n;
 48 double l,r;
 49 
 50 struct node {
 51     double x, y;
 52 }a[maxn];
 53 
 54 double dis(double x, int i) {
 55     return sqrt((a[i].y*a[i].y)+(a[i].x-x)*(a[i].x-x));
 56 }
 57 
 58 double Max(double a, double b) {
 59     if(b - a > eps) {
 60         return b;
 61     }
 62     return a;
 63 }
 64 
 65 double f(double x) {
 66     double u = 1, p = -1e8;
 67     for(int i = 1; i <= n; ++i) {
 68         p = Max(p,(a[i].y*a[i].y)+(a[i].x-x)*(a[i].x-x));
 69     }
 70     return sqrt(p);
 71 }
 72 
 73 double ts(double l, double r) {
 74     while(r >= l + eps) {
 75         //dbg(l),dbg(r);
 76         double lmid = l + (r-l)/3, rmid = r - (r-l)/3;
 77         if(f(lmid) <= f(rmid)) {
 78             r = rmid;
 79         }
 80         else {
 81             l = lmid;
 82         }
 83     }
 84     //dbg(l);
 85     return l;
 86 }
 87 
 88 int main()
 89 {
 90     memset(a, 0, sizeof(a));
 91     scanf("%d",&n);
 92 
 93     for(int i = 1; i <= n; ++i) {
 94         scanf("%lf %lf",&a[i].x, &a[i].y);
 95     }
 96     double mid = ts(-10000,10000);
 97     double ans = f(mid);
 98     printf("%.8lf\n",ans);
 99     return 0;
100 }

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