noi 2007 社交网络 最短路
题意：
大水题一道
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstring> 5 using namespace std; 6 #define MAXN 101 7 int map[MAXN][MAXN]; 8 long long path[MAXN][MAXN]; 9 int c[MAXN][MAXN][MAXN]; 10 double ans[MAXN]; 11 int n,m; 12 void solve() 13 { 14 memset(ans,0,sizeof(ans)); 15 int i,j,k; 16 for(k=1;k<=n;k++) 17 for(i=1;i<=n;i++) 18 for(j=1;j<=n;j++) 19 if(i!=j&&j!=k&&k!=i) 20 { 21 if(map[i][k]+map[k][j]==map[i][j]) 22 ans[k]+=(double)path[i][k]*path[k][j]/path[i][j]; 23 } 24 for(i=1;i<=n;i++) 25 printf("%.3lf\n",ans[i]); 26 } 27 28 void floyd() 29 { 30 int i,j,k; 31 for(k=1;k<=n;k++) 32 for(i=1;i<=n;i++) 33 for(j=1;j<=n;j++) 34 if(i!=j&&j!=k&&i!=k) 35 { 36 if(map[i][k]+map[k][j]<map[i][j]) 37 { 38 map[i][j]=map[i][k]+map[k][j]; 39 path[i][j]=path[i][k]*path[k][j]; 40 } 41 else if(map[i][k]+map[k][j]==map[i][j]) 42 path[i][j]+=path[i][k]*path[k][j]; 43 } 44 } 45 int main() 46 { 47 memset(path,0,sizeof(path)); 48 memset(map,0x3f,sizeof(map)); 49 int x,y,z; 50 scanf("%d%d",&n,&m); 51 int i; 52 for(i=1;i<=m;i++) 53 { 54 scanf("%d%d%d",&x,&y,&z); 55 map[x][y]=map[y][x]=z; 56 path[x][y]=path[y][x]=1; 57 } 58 floyd(); 59 solve(); 60 return 0; 61 }
