dfs 遍历 codeforces 24A

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//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 longusing namespace std;const int N = 110;int n, cur, ans;int g[N][N];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;}

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