BZOJ 1491 [NOI2007]社交网络

1491: [NOI2007]社交网络

Time Limit: 10 Sec Memory Limit: 64 MB
Submit: 1159 Solved: 660
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Description

Input

Output

输出文件包括n 行，每行一个实数，精确到小数点后3 位。第i 行的实数表示结点i 在社交网络中的重要程度。

Sample Input

4 4
1 2 1
2 3 1
3 4 1
4 1 1

Sample Output

1.000
1.000
1.000
1.000

HINT

为1

Source

题解：郑爷讲过，最简单的最短路计数了吧。

 1 /*=============================================================================
 2 * 
 3 *  Author: Crt_16
 4 * 
 5 *  Last modified: 2015-08-26 19:10
 6 * 
 7 *  Category:图论 最短路
 8 * 
 9 /*=============================================================================*/
10 
11 #include<iostream>
12 #include<cstdio>
13 #include<cmath>
14 #include<algorithm>
15 #include<stack>
16 #include<queue>
17 #include<cstring>
18 #define pau putchar(' ')
19 #define ent puts("");
20 #define mse(a,b) memset(a,b,sizeof(a))
21 #define ren(x) for(ted*e=fch[x];e;e=e->nxt)
22 #define rep(i,s,t) for(int i=s,__=t;i<=__;i++)
23 #define dwn(i,s,t) for(int i=s,__=t;i>=__;i--)
24 using namespace std;
25 const int maxn=100+10,maxm=4500+10,inf=1e9;
26 int d[maxn][maxn],n,m;long long f[maxn][maxn];
27 inline int read(){
28     int x=0;bool sig=true;char ch=getchar();
29     for(;!isdigit(ch);ch=getchar())if(ch=='-')sig=false;
30     for(;isdigit(ch);ch=getchar())x=10*x+ch-'0';return sig?x:-x;
31 }
32 inline void write(int x){
33     if(x==0){putchar('0');return;}if(x<0)putchar('-'),x=-x;
34     int len=0;static int buf[20];while(x)buf[len++]=x%10,x/=10;
35     for(int i=len-1;i>=0;i--)putchar(buf[i]+'0');return;
36 }
37 int main(){
38     n=read();m=read();int x,y;
39     rep(i,1,n)rep(j,1,n)d[i][j]=inf;rep(i,1,n)d[i][i]=0;
40     rep(i,1,m)x=read(),y=read(),d[x][y]=d[y][x]=read();
41     rep(i,1,n)rep(j,1,n)f[i][j]=1;
42     rep(k,1,n)rep(i,1,n)rep(j,1,n)if((i^j)&&(j^k)&&(k^i)){
43         if(d[i][j]>d[i][k]+d[k][j]){
44             d[i][j]=d[i][k]+d[k][j];
45             f[i][j]=f[i][k]*f[k][j];
46         }
47         else if(d[i][j]==d[i][k]+d[k][j])f[i][j]+=f[i][k]*f[k][j];
48     }
49     double ans[maxn];mse(ans,0.0);
50     rep(k,1,n)rep(i,1,n)rep(j,1,n)if((i^j)&&(j^k)&&(k^i)&&(d[i][j]==d[i][k]+d[k][j])){
51         ans[k]+=(double)f[i][k]*(double)f[k][j]/(double)f[i][j];
52     }rep(i,1,n)printf("%.3lf
",ans[i]);
53 
54     return 0;
55 }