Codeforces Round #236 (Div. 2)E. Strictly Positive Matrix(402E)

　　　　　　　　　　　　　　　　　　　　　　　　　　E. Strictly Positive Matrix

 

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 ton from 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).

Sample test(s)

input

2
1 0
0 1

output

NO

input

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： 矩阵matrix[n][n], 对角线上元素不全为0， 其他元素的值大于等于0. 问是否存在k 使得 矩阵的k次幂之后 元素的值全部大于0.

设 A = B²  , B 为一邻接矩阵， 则A[i][j] 的实际意义为 从i到j 经过一个点（不包含i, j）的路径的个数。。。对于k次幂就是经过k-1个点的路径的个数了。

 要使 A[i][j] 大于0 ，（有向图） i 到j必须连通。。 A[j][i] > 0 && A[i][j] < 0 ，则i, j之间必能形成回路。

可以理解为 从任意点开始都可以遍历整个有向图。

建两个图（正向 反向），分别跑dfs。。判断一下即可。 

#include <bits/stdc++.h>
using namespace std;
vector<int>G1[2015];
vector<int>G2[2015];
int  c1, c2;
bool vis[2015];
void dfs1(int r)
{
    c1++;
    vis[r] = true;
    for (int i = 0; i < G1[r].size(); i++)
    {
        if (!vis[G1[r][i]])
            dfs1(G1[r][i]);
    }
}
void dfs2(int r)
{
    c2++;
    vis[r] = true;
    for (int i = 0; i < G2[r].size(); i++)
    {
        if (!vis[G2[r][i]])
            dfs2(G2[r][i]);
    }
}
int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
    #endif
    int n;
    while (~scanf ("%d", &n))
    {
        for (int i = 0; i <= n; i++)
        {
            G1[i].clear();
            G2[i].clear();
        }
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                int x;
                scanf ("%d", &x);
                if (i != j && x)
                {
                    G1[i+1].push_back(j+1);
                    G2[j+1].push_back(i+1);
                }
            }
        }
        c1 = c2 = 0;
        memset(vis, false, sizeof(vis));
        dfs1(1);
        memset(vis, false, sizeof(vis));
        dfs2(1);
        printf("%s
", (c1 == n && c2 == n) ? "YES" : "NO");
    }
    return 0;
}