随机搜索与模拟退火
这是我见过的最神的乱搞!
开坑.
AC POJ 2420
1 #include <cstdio>
2 #include <fstream>
3 #include <iostream>
4
5 #include <cstdlib>
6 #include <cstring>
7 #include <algorithm>
8 #include <cmath>
9
10 #include <queue>
11 #include <vector>
12 #include <map>
13 #include <set>
14 #include <stack>
15 #include <list>
16
17 typedef unsigned int uint;
18 typedef long long int ll;
19 typedef unsigned long long int ull;
20 typedef double db;
21
22 using namespace std;
23
24 inline int getint()
25 {
26 int res=0;
27 char c=getchar();
28 bool mi=false;
29 while(c<'0' || c>'9') mi=(c=='-'),c=getchar();
30 while('0'<=c && c<='9') res=res*10+c-'0',c=getchar();
31 return mi ? -res : res;
32 }
33 inline ll getll()
34 {
35 ll res=0;
36 char c=getchar();
37 bool mi=false;
38 while(c<'0' || c>'9') mi=(c=='-'),c=getchar();
39 while('0'<=c && c<='9') res=res*10+c-'0',c=getchar();
40 return mi ? -res : res;
41 }
42
43 db eps=1e-20;
44 inline bool feq(db a,db b)
45 { return fabs(a-b)<eps; }
46
47 template<typename Type>
48 inline Type avg(const Type a,const Type b)
49 { return a+((b-a)/2); }
50
51
52 //==============================================================================
53 //==============================================================================
54 //==============================================================================
55 //==============================================================================
56
57
58
59 int n;
60
61 db x[105];
62 db y[105];
63
64 inline db dis2(db x1,db y1,db x2,db y2)
65 { return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); }
66
67 inline db getdist(db X,db Y)
68 {
69 db l=0;
70 for(int i=0;i<n;i++)
71 l+=dis2(X,Y,x[i],y[i]);
72 return l;
73 }
74
75 int _cnt=0;
76 int rands=0;
77 inline int getrand()
78 {
79 _cnt++;
80 if(_cnt==32768*32768) srand(++rands);
81 return rand()+32768*rand();
82 }
83
84 inline db dbrand()
85 { return (db)(rand()+32768*rand())/(db)(32768*32768); }
86
87 db res;
88
89
90 int main()
91 {
92 srand(rands);
93
94 while(scanf("%d",&n)>0)
95 {
96 for(int i=0;i<n;i++)
97 {
98 x[i]=getint();
99 y[i]=getint();
100 }
101
102 //Searching.
103 db cx=0,cy=0;
104 res=getdist(cx,cy);
105
106 db T=1e5;
107
108 while(T>=1e-2)
109 {
110 db rx=cx+(dbrand()-0.5)*T*2.0;
111 db ry=cy+(dbrand()-0.5)*T*2.0;
112 db dist=getdist(rx,ry);
113 db d=res-dist;
114 if(d>0)
115 {
116 res=dist;
117 cx=rx;
118 cy=ry;
119 }
120 T*=0.98;
121 }
122
123 printf("%.0f\n",res);
124 }
125
126
127 return 0;
128 }
注意程序写的并不是完整的模拟退火....我们只接受最优参数,而完全不接受那些比较差的参数......
但是这里的搜索范围使用了类似的温度变量T....
db rx=cx+(dbrand()-0.5)*T*2.0;
db ry=cy+(dbrand()-0.5)*T*2.0;
这应该也算模拟退火吧(模拟分子的不规则运动)........反正能A掉这个题........
AC POJ 1379
1 #include <cstdio>
2 #include <fstream>
3 #include <iostream>
4
5 #include <cstdlib>
6 #include <cstring>
7 #include <algorithm>
8 #include <cmath>
9
10 #include <queue>
11 #include <vector>
12 #include <map>
13 #include <set>
14 #include <stack>
15 #include <list>
16
17 typedef unsigned int uint;
18 typedef long long int ll;
19 typedef unsigned long long int ull;
20 typedef double db;
21
22 using namespace std;
23
24 inline int getint()
25 {
26 int res=0;
27 char c=getchar();
28 bool mi=false;
29 while(c<'0' || c>'9') mi=(c=='-'),c=getchar();
30 while('0'<=c && c<='9') res=res*10+c-'0',c=getchar();
31 return mi ? -res : res;
32 }
33 inline ll getll()
34 {
35 ll res=0;
36 char c=getchar();
37 bool mi=false;
38 while(c<'0' || c>'9') mi=(c=='-'),c=getchar();
39 while('0'<=c && c<='9') res=res*10+c-'0',c=getchar();
40 return mi ? -res : res;
41 }
42
43 db eps=1e-20;
44 inline bool feq(db a,db b)
45 { return fabs(a-b)<eps; }
46
47 template<typename Type>
48 inline Type avg(const Type a,const Type b)
49 { return a+((b-a)/2); }
50
51
52 //==============================================================================
53 //==============================================================================
54 //==============================================================================
55 //==============================================================================
56
57 int n;
58
59 db xlim,ylim;
60
61 db x[1050];
62 db y[1050];
63
64 inline db dis2(db x1,db y1,db x2,db y2)
65 { return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); }
66
67 inline db getdist(db X,db Y)
68 {
69 db l=1e40;
70 for(int i=0;i<n;i++)
71 l=min(l,dis2(X,Y,x[i],y[i]));
72 return l;
73 }
74
75 int _cnt=0;
76 int rands=0;
77 inline int getrand()
78 {
79 _cnt++;
80 if(_cnt==32768*32768) srand(++rands);
81 return rand()+32768*rand();
82 }
83
84 inline db dbrand()
85 { return (db)(rand()+32768*rand())/(db)(32768*32768); }
86
87 db res;
88
89 int main()
90 {
91 srand(rands);
92
93 int T=getint();
94 while(T--)
95 {
96 scanf("%lf%lf%d",&xlim,&ylim,&n);
97 for(int i=0;i<n;i++)
98 scanf("%lf%lf",x+i,y+i);
99
100 //Searching.
101 db cx=0,cy=0;
102 res=getdist(cx,cy);
103
104 db T=1e4;
105 while(T>=1e-4)
106 {
107 db rx=cx+(dbrand()-0.5)*2.0*T;
108 db ry=cy+(dbrand()-0.5)*2.0*T;
109
110 rx=max(0.0,rx); rx=min(xlim,rx);
111 ry=max(0.0,ry); ry=min(ylim,ry);
112
113 db dist=getdist(rx,ry);
114 if(dist>res)
115 {
116 res=dist;
117 cx=rx;
118 cy=ry;
119 }
120 T*=0.999;
121 }
122
123 printf("The safest point is (%.1f, %.1f).\n",cx,cy);
124 }
125
126 return 0;
127 }
跟上一题差不多,调一下参数加一点限制就交了.....然后就A了..... 313ms.....
依然是不带概率回退的参数搜索.......
T*0.999可以调整到T*0.998,这样以后总的运行时间是172ms...
AC BZOJ 3680
不带概率接受的模拟退火
1 #include <cstdio>
2 #include <fstream>
3 #include <iostream>
4
5 #include <cstdlib>
6 #include <cstring>
7 #include <algorithm>
8 #include <cmath>
9
10 #include <queue>
11 #include <vector>
12 #include <map>
13 #include <set>
14 #include <stack>
15 #include <list>
16
17 typedef unsigned int uint;
18 typedef long long int ll;
19 typedef unsigned long long int ull;
20 typedef double db;
21
22 using namespace std;
23
24 inline int getint()
25 {
26 int res=0;
27 char c=getchar();
28 bool mi=false;
29 while(c<'0' || c>'9') mi=(c=='-'),c=getchar();
30 while('0'<=c && c<='9') res=res*10+c-'0',c=getchar();
31 return mi ? -res : res;
32 }
33 inline ll getll()
34 {
35 ll res=0;
36 char c=getchar();
37 bool mi=false;
38 while(c<'0' || c>'9') mi=(c=='-'),c=getchar();
39 while('0'<=c && c<='9') res=res*10+c-'0',c=getchar();
40 return mi ? -res : res;
41 }
42
43 //==============================================================================
44 //==============================================================================
45 //==============================================================================
46 //==============================================================================
47
48
49 db frand()
50 { return db(rand()+rand()*32768)/(32768*32768-1); }
51
52 struct point
53 {
54 db x,y;
55 point(){ x=y=0; }
56 point(db x,db y):x(x),y(y){ }
57
58 point operator+(point f)
59 { return point(x+f.x,y+f.y); }
60
61 point operator-(point f)
62 { return point(x-f.x,y-f.y); }
63
64 point operator()(point f)
65 { return f-*this; }
66
67 point operator*(db k)
68 { return point(k*x,k*y); }
69
70 db len()
71 { return sqrt(x*x+y*y); }
72 };
73
74
75 int n;
76
77 point a[10050];
78 db w[10050];
79
80 db getdist(point x)
81 {
82 db res=0.0;
83 for(int i=0;i<n;i++)
84 res+=x(a[i]).len()*w[i];
85 return res;
86 }
87
88 int main()
89 {
90 srand(1782);
91
92 n=getint();
93 for(int i=0;i<n;i++)
94 scanf("%lf%lf%lf",&a[i].x,&a[i].y,&w[i]);
95
96 //simulate anneal
97 db B=1e10;
98 db T=B;
99 db R=0.992;
100 point res(0,0);
101 db dist=getdist(res);
102 while(T>=1e-4)
103 {
104 point cur(res.x+(frand()*2.0-1.0)*T,res.y+(frand()*2.0-1.0)*T);
105 db cd=getdist(cur);
106
107 if(cd<dist)
108 {
109 res=cur;
110 dist=cd;
111 }
112
113 T*=R;
114 }
115
116 printf("%.3f %.3f\n",res.x,res.y);
117
118 return 0;
119 }
加上了概率接受以后就再也没法A了TAT
而且好像T的初始值要设到非常非常大否则精度很难搞?
具体算法参考
http://www.cnblogs.com/heaad/archive/2010/12/20/1911614.html
