HDOJ 题目5097 Page Rank（矩阵运算，模拟）

Page Rank

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 100000/100000 K (Java/Others)

Total Submission(s): 280    Accepted Submission(s): 75

Problem Description

Evaluation and rank of web pages is a hot topic for many internet companies and researchers. PageRank is a link analysis tool and it assigns a numerical weighting to each element of a hyperlinked set of documents, such as the World Wide Web, with the purpose
of "measuring" its relative importance within the set. The algorithm may be applied to any collection of entities with reciprocal quotations and references. The numerical weight that it assigns to any given element E is referred to as the PageRank of E and
denoted by . Other factors like Author Rank can contribute to the importance of an entity.



For simplicity, in this problem PageRank vector q is defined as q = Gq, Where , 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 the example on the right, we have:








Denote the current PageRank vector and the next PageRank vector by q^cur and q^next respectively. The process is to compute the iterative powering for finding the first eigenvector.








The computation ends until  for some small ε(10^-10).

 

Input

The input contains many test cases. 



For each case, there are multiple lines. The first line contains an integer N(N<=3000), which represents the number of pages. Then a N*N zero-one matrix follows. The element E_ij (0 <= i, j < N) on the matrix represents whether the i-th page has a
hyper link to the j-th page.

 

Output

Output one line with the eigenvector. The numbers should be separated by a space and be correctly rounded to two decimal places.

 

Sample Input

4
0111
0011
0001
0100

 

Sample Output

0.15 1.49 0.83 1.53

 

Source

field=problem&key=2014%C9%CF%BA%A3%C8%AB%B9%FA%D1%FB%C7%EB%C8%FC%A1%AA%A1%AA%CC%E2%C4%BF%D6%D8%CF%D6%A3%A8%B8%D0%D0%BB%C9%CF%BA%A3%B4%F3%D1%A7%CC%E1%B9%A9%CC%E2%C4%BF%A3%A9&source=1&searchmode=source" style="color:rgb(26,92,200); text-decoration:none">2014上海全国邀请赛——题目重现（感谢上海大学提供题目）

 

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模拟，，，

add函数没有卵用。。

ac代码

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");
    }
}