2018 ACM-ICPC 南京站 D Country Meow

比眉伴天荒 2022-04-15 05:07 174阅读 0赞

## 题目：[点击打开链接][Link 1]   题意：最小球覆盖裸题。   分析：三分套三分再套三分（x,y,z三个方向嵌套三分）或者模拟退火。   代码一（三分）： ##

#include<algorithm>
    #include<iostream>
    #include<fstream>
    #include<complex>
    #include<cstdlib>
    #include<cstring>
    #include<cassert>
    #include<iomanip>
    #include<string>
    #include<cstdio>
    #include<bitset>
    #include<vector>
    #include<cctype>
    #include<cmath>
    #include<ctime>
    #include<stack>
    #include<queue>
    #include<deque>
    #include<list>
    #include<set>
    #include<map>
    using namespace std;
    #define pt(a) cout<<a<<endl
    #define debug test
    #define mst(ss,b) memset((ss),(b),sizeof(ss))
    #define rep(i,a,n) for (int i=a;i<=n;i++)
    #define per(i,a,n) for (int i=n-1;i>=a;i--)
    #define pii pair<int,int>
    #define fi first
    #define se second
    #define ll long long
    #define ull unsigned long long
    #define pb push_back
    #define mp make_pair
    #define inf 0x3f3f3f3f
    #define eps 1e-10
    #define PI acos(-1.0)
    const ll mod = 1e9+7;
    const int N = 1e2+10;
    
    ll qp(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
    int to[4][2]={
       {-1,0},{1,0},{0,-1},{0,1}};
    
    struct P{
        double x,y,z;
    }p[N];
    double dis(P A,P B){
        return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y)+(A.z-B.z)*(A.z-B.z));
    }
    int n;
    double cal(double x,double y,double z)
    {
        double sum=0;
        for(int i=0;i<n;i++)sum=max(sum,dis(p[i],(P){x,y,z}));
        return sum;
    }
    double cal(double x,double y)
    {
        double l=-100000,r=100000,ans=1e18;
        for(int i=1;i<=50;i++)
        {
            double mid=(l+r)/2;
            double m=(mid+r)/2;
            double L=cal(x,y,mid);
            double R=cal(x,y,m);
            if(L>R)l=mid;
            else r=m;
            ans=min(ans,L);
            ans=min(ans,R);
        }
        return ans;
    }
    double cal(double x)
    {
        double l=-100000,r=100000,ans=1e18;
        for(int i=1;i<=50;i++)
        {
            double mid=(l+r)/2;
            double m=(mid+r)/2;
            double L=cal(x,mid);
            double R=cal(x,m);
            if(L>R)l=mid;
            else r=m;
            ans=min(ans,L);
            ans=min(ans,R);
        }
        return ans;
    }
    int main()
    {
        cin>>n;
        for(int i=0;i<n;i++)scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].z);
        double l=-100000,r=100000,ans=1e18;
        for(int i=1;i<=50;i++)
        {
            double mid=(l+r)/2;
            double m=(mid+r)/2;
            double L=cal(mid);
            double R=cal(m);
            if(L>R)l=mid;
            else r=m;
            ans=min(ans,L);
            ans=min(ans,R);
        }
        printf("%.5f\n",ans);
        return 0;
    }

## 代码二（退火法）： ##

///#pragma comment(linker, "/STACK:102400000,102400000")
    ///#include<unordered_map>
    ///#include<unordered_set>
    #include<algorithm>
    #include<iostream>
    #include<fstream>
    #include<complex>
    #include<cstdlib>
    #include<cstring>
    #include<cassert>
    #include<iomanip>
    #include<string>
    #include<cstdio>
    #include<bitset>
    #include<vector>
    #include<cctype>
    #include<cmath>
    #include<ctime>
    #include<stack>
    #include<queue>
    #include<deque>
    #include<list>
    #include<set>
    #include<map>
    using namespace std;
    #define pt(a) cout<<a<<endl
    #define debug test
    #define mst(ss,b) memset((ss),(b),sizeof(ss))
    #define rep(i,a,n) for (int i=a;i<=n;i++)
    #define per(i,a,n) for (int i=n-1;i>=a;i--)
    #define pii pair<int,int>
    #define fi first
    #define se second
    #define ll long long
    #define ull unsigned long long
    #define pb push_back
    #define mp make_pair
    #define inf 0x3f3f3f3f
    #define eps 1e-10
    #define PI acos(-1.0)
    const ll mod = 1e9+7;
    const int N = 110;
    
    ll qp(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
    int to[4][2]={
       {-1,0},{1,0},{0,-1},{0,1}};
    
    int n;
    struct P{
        double x,y,z;
    }s,p[N];
    double delta,ans;
    
    double dis(P a,P b) {
        return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z));
    }
    
    void sa() {
        s.x=s.y=s.z=0;
        delta=10000,ans=1e20;
        ///delta开始要设大一点，不然会wa
        while(delta>eps) {
            int d=1;
            for(int i=2;i<=n;i++)
                if(dis(s,p[i])>dis(s,p[d])) d=i;
            double md=dis(s,p[d]);
            ans=min(ans,md);
            s.x+=(p[d].x-s.x)/md*delta;
            s.y+=(p[d].y-s.y)/md*delta;
            s.z+=(p[d].z-s.z)/md*delta;
            delta*=0.99;
        }
        cout<<fixed<<setprecision(7)<<ans<<endl;
    }
    
    int main() {
        ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
        while(cin>>n) {
            for(int i=1;i<=n;i++) cin>>p[i].x>>p[i].y>>p[i].z;
            sa();
        }
        return 0;
    }

[Link 1]: http://codeforces.com/gym/101981/attachments/download/7891/20182019-acmicpc-asia-nanjing-regional-contest-en.pdf