UVA 11090 – Going in Cycle!!

题目链接

题意：给定一个有向图，球平均权值最小的回路

思路：二分+判负环，每次二分一个值mid，判断是否存在小于mid的环，那么就是(w1 + w2 + w3…) / n < mid == w1 – mid + w2 – mid + w3 – mid …. < 0，所以每次二分的时候，把边权值减掉mid，之后bellmanford判负环即可

代码：

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;

typedef double Type;

const int MAXNODE = 55;

struct Edge {
	int u, v;
	Type dist;
	Edge() {}
	Edge(int u, int v, Type dist) {
		this->u = u;
		this->v = v;
		this->dist = dist;
	}
};

struct BellmanFrod {
	int n, m;
	vector<Edge> edges;
	vector<int> g[MAXNODE];
	bool inq[MAXNODE];
	Type d[MAXNODE];
	int p[MAXNODE];
	int cnt[MAXNODE];

	void init(int n) {
		this->n = n;
		for (int i = 0; i < n; i++) g[i].clear();
		edges.clear();
	}

	void add_Edge(int u, int v, Type dist) {
		edges.push_back(Edge(u, v, dist));
		m = edges.size();
		g[u].push_back(m - 1);
	}

	bool negativeCycle() {
		queue<int> Q;
		memset(inq, 0, sizeof(inq));
		memset(cnt, 0, sizeof(cnt));
		for (int i = 0; i < n; i++) {
			d[i] = 0; inq[i] = true; Q.push(i);
		}

		while (!Q.empty()) {
			int u = Q.front();
			Q.pop();
			inq[u] = false;
			for (int i = 0; i < g[u].size(); i++) {
				Edge& e = edges[g[u][i]];
				if (d[e.v] > d[u] + e.dist) {
					d[e.v] = d[u] + e.dist;
					p[e.v] = g[u][i];
					if (!inq[e.v]) {
						Q.push(e.v);
						inq[e.v] = true;
						if (++cnt[e.v] > n) return true;
					}
				}
			}
		}
		return false;
	}
} gao;

int T, n, m;

bool judge(double mid) {
	for (int i = 0; i < gao.m; i++)
		gao.edges[i].dist -= mid;
	bool tmp = gao.negativeCycle();
	for (int i = 0; i < gao.m; i++)
		gao.edges[i].dist += mid;
	return tmp;
}

int main() {
	scanf("%d", &T);
	int cas = 0;
	while (T--) {
		scanf("%d%d", &n, &m);
		gao.init(n);
		int u, v;
		double dist;
		double Max = 0;
		while (m--) {
			scanf("%d%d%lf", &u, &v, &dist);
			u--; v--;
			Max = max(Max, dist);
			gao.add_Edge(u, v, dist);
		}
		printf("Case #%d: ", ++cas);
		if (!judge(Max + 1)) printf("No cycle found.\n");
		else {
			double l = 0, r = Max;
			for (int i = 0; i < 100; i++) {
				double mid = (l + r) / 2;
				if (!judge(mid)) l = mid;
				else r = mid;
			}
			printf("%.2lf\n", l);
		}
	}
	return 0;
}