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题目分析 

        1.看数据范围，大概知道dfs能做

        2.自0问题开始查找，确保之后每次查找到的问题的困难度均大于上一次

        3.遍历所有情况再记录cnt即可

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代码

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
using ll = long long;
using ull = unsigned long long;
const int N = 20;

int g[N][N];
bool vis[N];
int n, ans;

void dfs(int x, int val, int cnt) {
    ans = (ans > cnt ? ans : cnt);
    for(int i = 0; i < n; i++)
    {
        if(vis[i] || i == x) continue; //走过或为当前点则continue
        if(g[x][i] >= val) 
        {
            vis[i] = 1;
            dfs(i, g[x][i], cnt + 1);
            vis[i] = 0;
        }
    }
}

int main()
{
    while(scanf("%d", &n) != EOF)
    {
        ans = 0;
        memset(vis, 0, sizeof vis);
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < n; j++)
            {
                scanf("%d", &g[i][j]);
            }
        }
        vis[0] = 1;
        dfs(0, 0, 1);
        printf("%d\n", ans);
    }
    return 0;
}