题意
每组数据读入一个n和n个字符串。定义前2个与末尾2个字母相同可以连接。问使这个环串的平均长度最大。求这个最大值。不存在输出No solution。思路
平均值公式:$$Average=(E_1+E_2+…..+E_n)/n$$ $$Average*n=(E_1+E_2+…+E_n)$$ $$(E_1-Average)+(E_2-Average)+…+(E_n-Average)=0$$ 那么可以二分答案:
$$(E_1-Ans)+(E_2-Ans)+…+(E_n-Ans)\geq0$$
然后瞎搞spfa。
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#include<algorithm> #include<bitset> #include<complex> #include<deque> #include<exception> #include<fstream> #include<functional> #include<iomanip> #include<ios> #include<iosfwd> #include<iostream> #include<istream> #include<iterator> #include<limits> #include<list> #include<locale> #include<map> #include<memory> #include<new> #include<numeric> #include<ostream> #include<queue> #include<set> #include<sstream> #include<stack> #include<stdexcept> #include<streambuf> #include<string> #include<typeinfo> #include<utility> #include<valarray> #include<vector> #include<cctype> #include<cerrno> #include<cfloat> #include<ciso646> #include<climits> #include<clocale> #include<cmath> #include<csetjmp> #include<csignal> #include<cstdarg> #include<cstddef> #include<cstdio> #include<cstdlib> #include<cstring> #include<ctime> #define E 200010 #define eps 1e-3 using namespace std; inline int read(){ int res=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();} while(ch>='0'&&ch<='9') res=res*10+ch-'0',ch=getchar(); return res*f; } inline void write(int x){ if(x<0) putchar('-'),x=-x; if(x<10) putchar(x+'0'); else{ write(x/10); putchar(x%10+'0'); } } //queue<int> q; //set<int> s; //priority_queue<int> q1; //priority_queue<int,vector<int>,greater<int> > q2; //list<int> l; //stack<int> s; int n; string str[100010]; int fir[E],nxt[E],son[E],tot,cnt,Max; double w[E],dis[E],flag; int vis[E]; int f[6666]; void add(int x,int y,double z){++tot;son[tot]=y;nxt[tot]=fir[x];fir[x]=tot;w[tot]=z;} void spfa(int s,int v,double mid){ if(flag==1) return ; vis[s]=v; for(int i=fir[s];i;i=nxt[i]){ int to=son[i]; if(dis[s]+w[i]>dis[to]+mid){ dis[to]=dis[s]+w[i]-mid; if(dis[to]>Max){ flag=1; return ; } if(!vis[to]) spfa(to,v,mid); if(flag) return ; else if(vis[to]==v){ flag=1; return ; } } } vis[s]=0; } bool check(double mid){ flag=0; for(int i=0;i<=cnt;i++) dis[i]=0.0; memset(vis,0,sizeof(vis)); for(int i=1;i<=cnt;i++){ spfa(i,i,mid); if(flag==1) break; } return flag; } int main(){ // freopen("code.in","r",stdin);freopen("code.out","w",stdout); n=read(); while(n!=0){ for(int i=1;i<=n;i++) cin>>str[i]; memset(fir,0,sizeof(fir)); memset(f,0,sizeof(f)); tot=0;cnt=0;Max=0; for(int i=1;i<=n;i++){ int len=str[i].length(); Max=max(Max,len); int a=(str[i][0]-'a')*26+str[i][1]-'a'; int b=(str[i][len-2]-'a')*26+str[i][len-1]-'a'; if(!f[a]) f[a]=++cnt; int A=f[a]; if(!f[b]) f[b]=++cnt; int B=f[b]; add(A,B,(double)len); } Max*=n; double l=0,r=1000,ans=-1,mid; while(l+eps<r){ mid=(l+r)/2; if(check(mid)) l=mid,ans=mid; else r=mid; } if(ans!=-1) printf("%.2lf\n",ans); else puts("No solution"); n=read(); } return 0; } |