E. Strictly Positive Matrix [强联通+矩阵幂转图论+缩点] 好题-白红宇

E. Strictly Positive Matrix [强联通+矩阵幂转图论+缩点] 好题

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发布时间：2019-03-08

本文共 2294 字，大约阅读时间需要 7 分钟。

You have matrix a of size n × n. Let's number the rows of the matrix from 1 to n from top to bottom, let's number the columns from 1 to nfrom left to right. Let's use aij to represent the element on the intersection of the i-th row and the j-th column.

Matrix a meets the following two conditions:

for any numbers i, j (1 ≤ i, j ≤ n) the following inequality holds: aij ≥ 0;

$\text{[math]}$ .

Matrix b is strictly positive, if for any numbers i, j (1 ≤ i, j ≤ n) the inequality bij > 0 holds. You task is to determine if there is such integer k ≥ 1, that matrix ak is strictly positive.

Input

The first line contains integer n (2 ≤ n ≤ 2000) — the number of rows and columns in matrix a.

The next n lines contain the description of the rows of matrix a. The i-th line contains n non-negative integers ai1, ai2, ..., ain(0 ≤ aij ≤ 50). It is guaranteed that $\text{[math]}$ .

Output

If there is a positive integer k ≥ 1, such that matrix ak is strictly positive, print "YES" (without the quotes). Otherwise, print "NO" (without the quotes).

          Examples 
            input 
             Copy 
21 00 1
            output 
NO
            input 
             Copy 
54 5 6 1 21 2 3 4 56 4 1 2 41 1 1 1 14 4 4 4 4
            output 
YES

题意:给定一个矩阵A,其中A[i][j]>=0 && sigma[i][i]>0 问,是否存在K,使得B=A^K,其中B[i][j]>0

思路:矩阵乘法转图论. A^k大于0,代表所有元素直接是联通.

B=A^K 中 b[i][j]代表i到j长度为k的路条数

#include
   
    #define bug cout <<"bug"<
    
     
       edge[MAX_N];int n,cur,cnt,top,mp[MAX_N],DFN[MAX_N],LOW[MAX_N],s[MAX_N];void tarjan(int u){    DFN[u]=LOW[u]=cur++;    s[++top]=u;    for(int i=0;i<(int)edge[u].size();i++){        int v=edge[u][i];        if(!DFN[v]){            tarjan(v);            LOW[u]=min(LOW[u],LOW[v]);        }        else if(DFN[v] && !mp[v])   LOW[u]=min(LOW[u],DFN[v]);    }    if(DFN[u]==LOW[u]){        int v=-1;        cnt++;        while(u!=v){            v=s[top--];            mp[v]=cnt;//            cout << v<<" ";        }//        puts("************");    }}int main(void){    read(n);    for(int i=1;i<=n;i++){        for(int j=1;j<=n;j++){            int x;read(x);            if(i==j)    continue;            if(x>0) edge[i].push_back(j);        }    }    for(int i=1;i<=n;i++)        if(!DFN[i]) tarjan(i);//    cout << cnt << endl;    if(cnt==1)  cout <<"YES"<

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