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

E. Strictly Positive Matrix time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

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;.

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 .

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 2 1 0 0 1 output NO input Copy 5 4 5 6 1 2 1 2 3 4 5 6 4 1 2 4 1 1 1 1 1 4 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<bits/stdc++.h> #define bug cout <<"bug"<<endl; #define read(x) scanf("%d",&x) using namespace std; typedef long long ll; const int MAX_N=2000+2; const int MOD=1e9+7; const int INF=0x3f3f3f3f; vector <int> 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"<<endl; else cout <<"NO"<<endl; return 0; }