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。

 /*=============================================================================
 *
 *  Author: Crt_16
 *
 *  Last modified: 2015-08-26 19:10
 *
 *  Category:图论 最短路
 *
 /*=============================================================================*/

 #include<iostream>
 #include<cstdio>
 #include<cmath>
 #include<algorithm>
 #include<stack>
 #include<queue>
 #include<cstring>
 #define pau putchar(' ')
 #define ent puts("");
 #define mse(a,b) memset(a,b,sizeof(a))
 #define ren(x) for(ted*e=fch[x];e;e=e->nxt)
 #define rep(i,s,t) for(int i=s,__=t;i<=__;i++)
 #define dwn(i,s,t) for(int i=s,__=t;i>=__;i--)
 using namespace std;
 const int maxn=+,maxm=+,inf=1e9;
 int d[maxn][maxn],n,m;long long f[maxn][maxn];
 inline int read(){
     int x=;bool sig=true;char ch=getchar();
     for(;!isdigit(ch);ch=getchar())if(ch=='-')sig=false;
     for(;isdigit(ch);ch=getchar())x=*x+ch-'';return sig?x:-x;
 }
 inline void write(int x){
     if(x==){putchar('');return;}if(x<)putchar('-'),x=-x;
     int len=;static int buf[];while(x)buf[len++]=x%,x/=;
     for(int i=len-;i>=;i--)putchar(buf[i]+'');return;
 }
 int main(){
     n=read();m=read();int x,y;
     rep(i,,n)rep(j,,n)d[i][j]=inf;rep(i,,n)d[i][i]=;
     rep(i,,m)x=read(),y=read(),d[x][y]=d[y][x]=read();
     rep(i,,n)rep(j,,n)f[i][j]=;
     rep(k,,n)rep(i,,n)rep(j,,n)if((i^j)&&(j^k)&&(k^i)){
         if(d[i][j]>d[i][k]+d[k][j]){
             d[i][j]=d[i][k]+d[k][j];
             f[i][j]=f[i][k]*f[k][j];
         }
         else if(d[i][j]==d[i][k]+d[k][j])f[i][j]+=f[i][k]*f[k][j];
     }
     double ans[maxn];mse(ans,0.0);
     rep(k,,n)rep(i,,n)rep(j,,n)if((i^j)&&(j^k)&&(k^i)&&(d[i][j]==d[i][k]+d[k][j])){
         ans[k]+=(double)f[i][k]*(double)f[k][j]/(double)f[i][j];
     }rep(i,,n)printf("%.3lf\n",ans[i]);

     return ;
 }