• 模拟退火


     1 #include<cstdio> 
     2 #include<cstring>
     3 #include<cmath>
     4 #include<ctime>
     5 #include<iostream>
     6 #include<algorithm>
     7 #include<queue>
     8 #include<set>
     9 #define inf (0x7fffffff)
    10 #define l(a) ((a)<<1)
    11 #define r(a) ((a)<<1|1)
    12 #define b(a) (1<<(a))
    13 #define T_min (1e-8)
    14 #define R (0.99999)
    15 #define e (2.718281828459045)
    16 #define pi (3.14159265358979323846)
    17 #define rep(i,a,b) for(int i=a;i<=(b);i++)
    18 #define clr(a) memset(a,0,sizeof(a))
    19 typedef long long ll;
    20 typedef unsigned long long ull;
    21 using namespace std;
    22 int readint(){
    23     int t=0,f=1;char c=getchar();
    24     while(!isdigit(c)){
    25         if(c=='-') f=-1;
    26         c=getchar();
    27     }
    28     while(isdigit(c)){
    29         t=(t<<3)+(t<<1)+c-'0';
    30         c=getchar();
    31     }
    32     return t*f;
    33 }
    34 const int maxn=200009;
    35 int n,opt,tmp,pre,del,path[2][maxn];
    36 double T=1e5,d[maxn];
    37 struct node{
    38     double x,y;
    39     inline bool operator<(const node A)const{
    40         return x<A.x||(x==A.x&&y<A.y);
    41     }
    42 }X[maxn];
    43 int dis(int i,int j){
    44      return int(sqrt((X[i].x-X[j].x)*(X[i].x-X[j].x)+(X[i].y-X[j].y)*(X[i].y-X[j].y))); 
    45 }
    46 int f(int t){
    47     int s=0;
    48     rep(i,2,n+1) s+=dis(path[t][i-1],path[t][i]);
    49     return s;
    50 }
    51 void newway(){
    52     int type=rand()%3,a=rand()%(n-1),b=rand()%(n-1);
    53     while(a==b) a=rand()%(n-1);a+=2;b+=2;
    54     if(a>b) swap(a,b);
    55     if(type==0){
    56         rep(i,1,n+1) path[opt^1][i]=path[opt][i];
    57         swap(path[opt^1][a],path[opt^1][b]);
    58     }else if(type==1){
    59         rep(i,1,a-1) path[opt^1][i]=path[opt][i];
    60         rep(i,a,b) path[opt^1][i]=path[opt][a-i+b];
    61         rep(i,b+1,n+1) path[opt^1][i]=path[opt][i];
    62     }else{
    63         rep(i,1,a-1) path[opt^1][i]=path[opt][i];
    64         path[opt^1][a]=path[opt][b];
    65         rep(i,a+1,b) path[opt^1][i]=path[opt][i-1];
    66         rep(i,b+1,n+1) path[opt^1][i]=path[opt][i];
    67     }
    68 }
    69 void cal(){
    70     int ans,cnt=0;
    71     srand(time(NULL));
    72     rep(i,1,n) path[opt][i]=i;path[opt][n+1]=1;ans=pre=f(opt);
    73     while(T>T_min){
    74         cnt++;if(cnt>1e6) break;
    75         newway();
    76         tmp=f(opt^1);del=tmp-pre;
    77         ans=min(ans,tmp);
    78         if(del<0){
    79             opt^=1;pre=tmp;
    80         }else if(rand()*rand()/32767.0/32767.0<pow(e,-del/T)){
    81             opt^=1;pre=tmp;
    82         }
    83         T*=R;
    84     }
    85     cout<<cnt<<endl;
    86     printf("%d
    ",ans);
    87 }
    88 int main(){
    89     freopen("#input.txt","r",stdin);
    90     //freopen("#output.txt","w",stdout);
    91     n=readint();
    92     rep(i,1,n){
    93         X[i].x=readint();X[i].y=readint();
    94     }
    95     cal();
    96     fclose(stdin);
    97     //fclose(stdout);
    98     return 0;
    99 }
    TSP
  • 相关阅读:
    Tool工具页面代码
    Tool工具生成代码数据库Model生成代码
    类别切换 分页
    ASP.NET AJAX无刷新验证用户名
    VSS的配置和使用
    js 常用方法大全
    灵异——1995年北京330路公交车失踪案
    C#用HttpWebRequest通过代理服务器验证后抓取网页内容 。。。。。
    win2003远程 客户端无法连接到远程计算机。
    .net中点击button按钮显示下一条记录(上一条 下一条)
  • 原文地址:https://www.cnblogs.com/chensiang/p/7892802.html
Copyright © 2020-2023  润新知