codeforces679 D. Bear and Chase
codeforces679 D. Bear and Chase
http://blog.csdn.net/Ezereal/article/details/52743686
D Bear and Chase
O(n^3) 暴力
首先 ，枚举第一次放BCD地方
然后枚举罪犯离BCD的距离
然后枚举一下第二次方BCD的地方
再加一点预处理就行了
1 #include<cstdio> 2 #include<cstring> 3 using namespace std; 4 const int N=440; 5 const double eps=1e-9; 6 int i,j,k,l,n,m,ch,En,x,y; 7 int h[N],du[N],a[N][N],b[N][N],nn[N]; 8 double ans,now,p[N],q[N]; 9 struct cc { double p;int x;} A[N]; 10 struct edge { int s,n;} E[N*N]; 11 void R(int &x) { 12 x=0;ch=getchar(); 13 while (ch<'0' || '9'<ch) ch=getchar(); 14 while ('0'<=ch && ch<='9') x=x*10+ch-'0',ch=getchar(); 15 } 16 void E_add(int x,int y) { 17 du[x]++;du[y]++; 18 a[x][y]=a[y][x]=1; 19 E[++En].s=y;E[En].n=h[x];h[x]=En; 20 E[++En].s=x;E[En].n=h[y];h[y]=En; 21 } 22 double max(double x,double y) { 23 if (x>y) return x; 24 return y; 25 } 26 double get_num(int x) { 27 int i,j,k,nm=0; 28 for (i=1;i<=n;i++) p[i]=0.0; 29 for (i=1;i<=nn[x];i++) 30 for (k=h[b[x][i]];k;k=E[k].n) p[E[k].s]+=1.0/(1.0*n*du[b[x][i]]); 31 for (i=1;i<=n;i++) if (p[i]>eps) { 32 A[++nm].p=p[i]; 33 A[nm].x=i; 34 } 35 double ans=0.0,now; 36 for (i=1;i<=n;i++) { 37 now=0.0; 38 for (j=1;j<=nm;j++) q[a[i][A[j].x]]=0.0; 39 for (j=1;j<=nm;j++) if (A[j].p>q[a[i][A[j].x]]) { 40 now+=A[j].p-q[a[i][A[j].x]]; 41 q[a[i][A[j].x]]=A[j].p; 42 } 43 ans=max(ans,now); 44 } 45 return ans; 46 } 47 int main() { 48 memset(a,60,sizeof a); 49 R(n);R(m); 50 for (i=1;i<=n;i++) a[i][i]=0; 51 for (i=1;i<=m;i++) { 52 R(x);R(y); 53 E_add(x,y); 54 } 55 for (k=1;k<=n;k++) 56 for (i=1;i<=n;i++) 57 for (j=1;j<=n;j++) if (a[i][k]+a[k][j]<a[i][j]) a[i][j]=a[i][k]+a[k][j]; 58 for (i=1;i<=n;i++) { 59 now=0.0; 60 for (j=0;j<n;j++) nn[j]=0; 61 for (j=1;j<=n;j++) { 62 k=a[i][j]; 63 b[k][++nn[k]]=j; 64 } 65 for (j=0;j<n;j++) if (nn[j]) { 66 if (nn[j]==1) now+=1.0/(1.0*n); 67 else now+=max(1.0/(1.0*n),get_num(j)); 68 } 69 ans=max(ans,now); 70 } 71 printf("%.9lf\n",ans); 72 }
