模拟退火——模板
題目背景為TSP。
?
View Code 1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 #include<ctime> 7 8 using namespace std; 9 10 const int maxn=100; 11 const double max_t=10000.0; 12 const double ratio=0.98; 13 const double eps=1e-6; 14 const int per_t=25; 15 const int tT=1; 16 17 int n; 18 19 int ans; 20 21 bool use[maxn]; 22 23 int map[maxn][maxn]; 24 25 int z[maxn],y[maxn]; 26 27 void calc() 28 { 29 for (int T=1;T<=tT;T++) 30 { 31 memset(use,false,sizeof(use)); 32 use[0]=true; 33 long long nowans=0; 34 use[1]=true; 35 z[1]=1; 36 for (int a=2;a<=n;a++) 37 { 38 int wx=0; 39 while (use[wx]) 40 wx=rand()%n+1; 41 z[a]=wx; 42 use[wx]=true; 43 nowans+=map[z[a-1]][z[a]]; 44 } 45 nowans+=map[z[n]][z[1]]; 46 for (double nowt=max_t;nowt>=eps;nowt*=ratio) 47 { 48 for (int hehe=1;hehe<=per_t;hehe++) 49 { 50 int nowp1=rand()%(n-1); 51 int nowp2=rand()%(n-1); 52 while (nowp1==nowp2) 53 { 54 nowp1=rand()%(n-1); 55 nowp2=rand()%(n-1); 56 } 57 nowp1+=2;nowp2+=2; 58 if (nowp1>nowp2) swap(nowp1,nowp2); 59 long long newans=0; 60 swap(z[nowp1],z[nowp2]); 61 for (int a=1;a<n;a++) 62 newans+=map[z[a]][z[a+1]]; 63 newans+=map[z[n]][z[1]]; 64 long long delta=nowans-newans; 65 if (newans<nowans || exp((double)delta/nowt)>rand()/RAND_MAX) 66 { 67 nowans=newans; 68 } 69 else swap(z[nowp1],z[nowp2]); 70 if (ans>nowans) 71 { 72 for (int a=1;a<=n;a++) 73 y[a]=z[a]; 74 ans=nowans; 75 } 76 } 77 } 78 } 79 } 80 81 int main() 82 { 83 scanf("%d",&n); 84 if (n==1) 85 { 86 printf("0\n"); 87 return 0; 88 } 89 for (int a=1;a<=n;a++) 90 for (int b=1;b<=n;b++) 91 scanf("%d",&map[a][b]); 92 ans=123456789; 93 srand(time(0)); 94 calc(); 95 printf("%d\n",ans); 96 for (int a=1;a<=n;a++) 97 { 98 printf("%d",y[a]-1); 99 if (a==n) printf("\n"); 100 else printf(" "); 101 } 102 103 return 0; 104 }?
?
?
轉載于:https://www.cnblogs.com/zhonghaoxi/archive/2012/07/10/2584347.html
總結
- 上一篇: SQLServer转义问题
- 下一篇: 【原】Unity3D 窗口裁剪