P1522 牛的旅行 Cow Tours(floyd)
给你平面上的一些点,一开始有一些点是相连的,相连的边权是两点间的欧几里得距离,输入数据保证至少存在一组互相不相连的点
把相连的点叫做一个联通块,一个联通块的直径是其中任意两点间最短路长度的最大值
现在你需要在两个联通块中间连一条边,使得新生成的联通块的直径最小
一开始胡了个复杂度爆炸的算法上去,O(n^4),先跑一边floyd,然后暴力枚举在哪两个点连边,每次n^2更新距离并且找距离最大值,60分
看了题解才发现我是憨憨
首先用并查集维护联通块,对每个点用n^2时间求出在当前联通块中距离它最远的点到他的距离,设为mx,并且还要维护当前联通块的直径
然后n^2暴力枚举连边,假设在i、j之间连一条边,那么答案更新为:max(mx[i]+mx[j]+i、j的距离,max(i联通块的直径,j联通块的直径))
我被自己代码模板里int的max坑惨了,怪不得输出的时候丢精度
代码:
#include <bits/stdc++.h> #define int long long #define sc(a) scanf("%lld",&a) #define scc(a,b) scanf("%lld %lld",&a,&b) #define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c) #define schar(a) scanf("%c",&a) #define pr(a) printf("%lld",a) #define fo(i,a,b) for(int i=a;i<b;++i) #define re(i,a,b) for(int i=a;i<=b;++i) #define rfo(i,a,b) for(int i=a;i>b;--i) #define rre(i,a,b) for(int i=a;i>=b;--i) #define prn() printf("\n") #define prs() printf(" ") #define mkp make_pair #define pii pair<int,int> #define pub(a) push_back(a) #define pob() pop_back() #define puf(a) push_front(a) #define pof() pop_front() #define fst first #define snd second #define frt front() #define bak back() #define mem0(a) memset(a,0,sizeof(a)) #define memmx(a) memset(a,0x3f3f,sizeof(a)) #define memmn(a) memset(a,-0x3f3f,sizeof(a)) #define debug #define db double #define yyes cout<<"YES"<<endl; #define nno cout<<"NO"<<endl; using namespace std; typedef vector<int> vei; typedef vector<pii> vep; typedef map<int,int> mpii; typedef map<char,int> mpci; typedef map<string,int> mpsi; typedef deque<int> deqi; typedef deque<char> deqc; typedef priority_queue<int> mxpq; typedef priority_queue<int,vector<int>,greater<int> > mnpq; typedef priority_queue<pii> mxpqii; typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii; const int maxn=500005; const int inf=0x3f3f3f3f3f3f3f3f; const int MOD=100000007; const db eps=1e-10; int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;} int lowbit(int x){return x&-x;} int max(int a,int b){return a>b?a:b;} int min(int a,int b){return a<b?a:b;} int mmax(int a,int b,int c){return max(a,max(b,c));} int mmin(int a,int b,int c){return min(a,min(b,c));} void mod(int &a){a+=MOD;a%=MOD;} bool chk(int now){} int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;} int ll(int p){return p<<1;} int rr(int p){return p<<1|1;} int mm(int l,int r){return (l+r)/2;} int lg(int x){if(x==0) return 1;return (int)log2(x)+1;} bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;} db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));} bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;} inline int read(){ char ch=getchar();int s=0,w=1; while(ch<48||ch>57){if(ch=='-')w=-1;ch=getchar();} while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();} return s*w; } inline void write(int x){ if(x<0)putchar('-'),x=-x; if(x>9)write(x/10); putchar(x%10+48); } int n,p[155]; db x[155],y[155]; char s[155]; char c[155][155]; db dis[155][155],mx[155],big[155]; void floyd(){ re(k,1,n){ re(i,1,n){ re(j,1,n){ dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]); } } } } int _find(int x){ if(x!=p[x]) p[x]=_find(p[x]); return p[x]; } void _merge(int x,int y){ x=_find(x); y=_find(y); if(x!=y) p[x]=y; } signed main(){ ios_base::sync_with_stdio(0); cin.tie(0),cout.tie(0); sc(n); re(i,1,n) p[i]=i; re(i,1,n) scanf("%lf%lf",&x[i],&y[i]); re(i,1,n){ scanf("%s",s+1); re(j,1,n) c[i][j]=s[j]; } re(i,1,n){ re(j,1,n){ if(i==j) dis[i][j]=0.0; else{ if(c[i][j]=='1') dis[i][j]=len(x[i],y[i],x[j],y[j]),_merge(i,j); else dis[i][j]=inf; } } } db ans=inf; floyd(); memmn(mx); memmn(big); re(i,1,n){ re(j,1,n){ if(_find(i)!=_find(j)||dis[i][j]==inf) continue; mx[i]=max(mx[i],dis[i][j]); big[_find(i)]=max(big[_find(i)],mx[i]); } } re(i,1,n){ re(j,1,n){ if(_find(i)==_find(j)) continue; db tmp=mx[i]+mx[j]+len(x[i],y[i],x[j],y[j]); ans=min(ans,max(tmp,max(big[_find(i)],big[_find(j)]))); } } printf("%.6lf",ans); return 0; }