HDU 3007 模擬退火演算法
阿新 • • 發佈:2018-11-11
Buried memory
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4067 Accepted Submission(s): 2171
The world king Sconbin is not the exception.One day,Sconbin was sleeping,then swakened by one nightmare.It turned out that his love letters to Dufein were made public in his dream.These foolish letters might ruin his throne.Sconbin decided to destroy the letters by the military exercises's opportunity.The missile is the best weapon.Considered the execution of the missile,Sconbin chose to use one missile with the minimum destruction.
Sconbin had writen N letters to Dufein, she buried these letters on different places.Sconbin got the places by difficult,he wants to know where is the best place launch the missile,and the smallest radius of the burst area. Let's help Sconbin to get the award.
Sample Output 2.00 2.00 1.41
做法類似POJ2069
請參考:https://www.cnblogs.com/qq965921539/p/9806603.html
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <string> 5 #include <map> 6 #include <set> 7 #include <list> 8 #include <deque> 9 #include <queue> 10 #include <stack> 11 #include <vector> 12 #include <cmath> 13 #include <algorithm> 14 using namespace std; 15 #define it iterator 16 #define ll long long 17 #define eb emplace_back 18 #define lowbit(x) x & -x 19 #define all(x) x.begin(),x.end() 20 #define ZERO(a) memset(a,0,sizeof(a)) 21 #define MINUS(a) memset(a,0xff,sizeof(a)) 22 #define per(x,a,b) for(int x = a; x <= b; x++) 23 #define rep(x,a,b) for(int x = a; x >= b; x--) 24 #define IO ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) 25 26 const int birth = 19260817; 27 const int mo = 998244353; 28 const int maxn = 1e5 + 10; 29 const int mod = 1e9 + 7; 30 const int INF = 0x3fffffff; 31 const double eps = 1e-8; 32 33 //******************THE PROGRAM BEGINING****************** 34 struct node 35 { 36 double x, y; 37 }p[510],now; 38 39 double dis(node a, node b) 40 { 41 return sqrt(pow(a.x - b.x, 2) + pow(a.y - b.y, 2)); 42 } 43 44 double solve(int n) 45 { 46 double ans, cmp; 47 double T = 100.0; 48 double delat = 0.98; 49 now.x = now.y = 0.0; 50 int pos = 0; 51 while (T > eps) 52 { 53 pos = 0; 54 ans = dis(now, p[pos]); 55 per(i, 0, n - 1) 56 { 57 cmp = dis(now, p[i]); 58 if (cmp > ans) 59 { 60 pos = i; 61 ans = cmp; 62 } 63 } 64 now.x += (p[pos].x - now.x) / ans * T; 65 now.y += (p[pos].y - now.y) / ans * T; 66 T *= delat; 67 } 68 return ans; 69 } 70 71 int main() 72 { 73 int n; 74 while (scanf("%d", &n) && n) 75 { 76 per(i, 0, n - 1) 77 scanf("%lf %lf", &p[i].x, &p[i].y); 78 double ans = solve(n); 79 printf("%.2lf %.2lf %.2lf\n", now.x, now.y, ans); 80 } 81 return 0; 82 }