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
21 00 1
output
NO
input
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： 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。
1 #include <bits/stdc++.h>
2 using namespace std;
3 vector<int>G1[2015];
4 vector<int>G2[2015];
5 int c1, c2;
6 bool vis[2015];
7 void dfs1(int r)
8 {
9 c1++;
10 vis[r] = true;
11 for (int i = 0; i < G1[r].size(); i++)
12 {
13 if (!vis[G1[r][i]])
14 dfs1(G1[r][i]);
15 }
16 }
17 void dfs2(int r)
18 {
19 c2++;
20 vis[r] = true;
21 for (int i = 0; i < G2[r].size(); i++)
22 {
23 if (!vis[G2[r][i]])
24 dfs2(G2[r][i]);
25 }
26 }
27 int main()
28 {
29 #ifndef ONLINE_JUDGE
30 freopen("in.txt","r",stdin);
31 #endif
32 int n;
33 while (~scanf ("%d", &n))
34 {
35 for (int i = 0; i <= n; i++)
36 {
37 G1[i].clear();
38 G2[i].clear();
39 }
40 for (int i = 0; i < n; i++)
41 {
42 for (int j = 0; j < n; j++)
43 {
44 int x;
45 scanf ("%d", &x);
46 if (i != j && x)
47 {
48 G1[i+1].push_back(j+1);
49 G2[j+1].push_back(i+1);
50 }
51 }
52 }
53 c1 = c2 = 0;
54 memset(vis, false, sizeof(vis));
55 dfs1(1);
56 memset(vis, false, sizeof(vis));
57 dfs2(1);
58 printf("%s\n", (c1 == n && c2 == n) ? "YES" : "NO");
59 }
60 return 0;
61 }