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, S is the destination-by-source stochastic matrix, U is all one matrix, n is the number of
nodes and α is the weight between 0 and 1 (here we use 0.85).
for some small ε(10-10).4 0111 0011 0001 0100
0.15 1.49 0.83 1.53
Problem : 5097 ( Page Rank ) Judge Status : Accepted
RunId : 14492913 Language : C++ Author : lwj1994
Code Render Status : Rendered By HDOJ C++ Code Render Version 0.01 Beta
#include<stdio.h>
#include<string.h>
#include<math.h>
#define eps 1e-10
char map[3030];
double ans[3030][3030];
double q[2][3030];
int n;
void muti(double a[][3030],double num)
{
int i,j;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
a[i][j]*=num;
}
}
void add(double a[][3030],double b[][3030])
{
int i,j;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
a[i][j]+=b[i][j];
}
}
int jud()
{
double ans=0;
int i;
for(i=0;i<n;i++)
{
ans+=(q[0][i]-q[1][i])*(q[0][i]-q[1][i]);
}
//ans=sqrt(ans);
if(fabs(ans)<eps)
return 1;
return 0;
}
int main()
{
//int n;
while(scanf("%d",&n)!=EOF)
{
int i,j;
double a=0.85;
memset(ans,0,sizeof(ans));
for(i=0;i<n;i++)
{
int sum=0;
scanf("%s",&map);
for(j=0;j<n;j++)
{
if(map[j]=='1')
sum++;
//U[i][j]=1;
}
for(j=0;j<n;j++)
{
if(map[j]=='1')
ans[j][i]=1.0/sum;
}
}
muti(ans,a);
// muti(U,1.0/n*(0.15));
// printf("\n");
// add(ans,U);
for(i=0;i<n;i++)
for(j=0;j<n;j++)
ans[i][j]+=(0.15/n);
int p=0;
for(i=0;i<n;i++)
{
q[p][i]=1;
q[p^1][i]=0;
}
//mmuti(q[p],ans,q[p^1]);
//p^=1;
while(!jud())
{
//q[p^1]=mmuti(q[p],ans);
for(i=0;i<n;i++)
{
q[p^1][i]=0;
for(j=0;j<n;j++)
q[p^1][i]+=q[p][j]*ans[i][j];
}
p^=1;
}
printf("%.2lf",q[p][0]);
for(i=1;i<n;i++)
{
printf(" %.2lf",q[p][i]);
}
printf("\n");
}
}版权声明：本文为博主原创文章，未经博主允许不得转载。
HDOJ 题目5097 Page Rank（矩阵运算，模拟）
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原文地址：http://blog.csdn.net/yu_ch_sh/article/details/47612997
