HDOJ 题目5097 Page Rank（矩阵运算，模拟）

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标签：content each add tween run cti between man actor

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 技术分享

, 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:

技术分享

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.

技术分享

The computation ends until 技术分享

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

Sample Output

0.15 1.49 0.83 1.53

Source

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");
    }
}

HDOJ 题目5097 Page Rank（矩阵运算，模拟）

标签：content each add tween run cti between man actor

原文地址：http://www.cnblogs.com/mfmdaoyou/p/6882868.html

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