dfs 遍历 codeforces 24A

//题目中明确说了，这是一个环，要充分利用性质

//答案一定是在从1出发的两条路径中的一条，如果当前边反向，则更新当前值，最后答案就是两条路径中小的

//dfs时，记录父亲节点，这样就可以避免在向下一步走的时候，又回到了父亲

/*************************************************************************
> File Name: 24A.cpp
> Author: flyasdfvcxz
> Mail: 1099431883@qq.com
> Created Time: Fri 07 Nov 2014 08:10:08 PM CST
************************************************************************/

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <queue>
#include <map>
#include <ctime>
#include <cmath>
#define ll long long
using namespace std;
const int N = 110;
int n, cur, ans;
int g

;
void dfs(int x, int sum, int fa)
{
//cout << " x = " << x << " sum = " << sum << endl;
if (x == 1)
{
++cur;
if (cur > 1)
{
ans = min(ans, sum);
return;
}
}
for (int i = 1; i <= n; ++i)
{
if (i == fa) continue;
if (g[x][i])
{
dfs(i, sum, x);
}
else if (g[i][x])
{
dfs(i, sum + g[i][x], x);
}
}
}
int main()
{
//freopen("in.txt", "r", stdin);
while (cin >> n)
{
memset(g, 0, sizeof(g));
for (int i = 1; i <= n; ++i)
{
int a, b, c;
cin >> a >> b >> c;
g[a][b] = c;
}
ans = 1e9;
cur = 0;
dfs(1, 0, -1);
cout << ans << endl;
}
return 0;
}