[CF24A]Ring road(2019-11-15考试)

题目大意

给你一个\(n\)个点的环，每条边有方向，改变第\(i\)条边的方向代价为\(w_i\)，问将其改为强连通图的最小代价。\(n\leqslant100\)

题解

求出把边全部改为顺时针和全部改为逆时针的代价，较少的输出即可

卡点

无

C++ Code：

#include <cstdio>
#include <iostream>
#include <algorithm>
const int maxn = 111;

int head[maxn], cnt;
struct Edge { int to, nxt; } e[maxn << 1];
void addedge(int a, int b) {
	e[++cnt] = (Edge) { b, head[a] }; head[a] = cnt;
	e[++cnt] = (Edge) { a, head[b] }; head[b] = cnt;
}

int n, dep[maxn], l[maxn], r[maxn], w[maxn], res1, res2;
void dfs(int u) {
	for (int i = head[u], v; i; i = e[i].nxt) {
		v = e[i].to;
		if (!dep[v]) dep[v] = dep[u] + 1, dfs(v);
	}
}

int main() {
	std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0);
	std::cin >> n;
	for (int i = 1; i <= n; ++i) {
		std::cin >> l[i] >> r[i] >> w[i];
		addedge(l[i], r[i]);
	}
	dfs(dep[1] = 1);
	for (int i = 1; i <= n; ++i) {
		if (dep[r[i]] != dep[l[i]] % n + 1) res2 += w[i];
		else res1 += w[i];
	}
	std::cout << std::min(res1, res2) << '\n';
	return 0;
}