UVA - 11090 Going in Cycle!! (Bellman-Ford算法判负环)

Description

I IU P C2 0 06

Problem G: Going in Cycle!!

Input: standard input

Output: standard output

 

You are given a weighted directed graph with n vertices and m edges. Each cycle in the graph has a weight, which equals to sum of its edges. There are so many cycles in the graph with different weights. In this problem we want to find a cycle with the minimum mean.

 

Input

The first line of input gives the number of cases, N. N test cases follow. Each one starts with two numbers n andm. m lines follow, each has three positive numbera, b, c which means there is an edge from vertex a tob with weight of c.

 

Output

For each test case output one line containing “Case #x: ” followed by a number that is the lowest mean cycle in graph with 2 digits after decimal place, if there is a cycle. Otherwise print “No cycle found.”.

 

Constraints

-           n ≤ 50

-           a, b ≤ n

-           c ≤ 10000000

 

Sample Input

Output for Sample Input

2
2 1
1 2 1
2 2
1 2 2
2 1 3

Case #1: No cycle found.
Case #2: 2.50

题意：给定一个n个点m条边的加权有向图，求平均权值最小的回路。

思路：先使用二分求解mid。假设存在一个包含k条边的回路，回路上各个边的权值为w1,w2....wk,那么平均值小于mid

意味着：w1+w2..+wk<K*mid -> (w1-mid)+(w2-mid)+...(wk-mid) < 0，就是判断新图中是否含有负权回路了，两种初始化结果都对

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <queue>#include <vector>using namespace std;const int maxn = 60;const int inf = 0x3f3f3f3f;struct Edge {int from, to;double dist;};struct BellmanFord {int n, m;vector<Edge> edges;vector<int> G[maxn];bool inq[maxn];double d[maxn];int p[maxn];int cnt[maxn];void init(int n) {this->n = n;for (int i = 0; i < n; i++)G[i].clear();edges.clear();}void AddEdge(int from, int to, int dist) {edges.push_back((Edge){from, to, dist});m = edges.size();G[from].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[0] = 1;Q.push(i);}*/for (int i = 0; i < n; i++) {d[i] = 0;inq[i] = 1;Q.push(i);cnt[i] = 1;}while (!Q.empty()) {int u = Q.front();Q.pop();inq[u] = 0;for (int i = 0; i < G[u].size(); i++) {Edge &e = edges[G[u][i]];if (d[e.to] > d[u] + e.dist) {d[e.to] = d[u] + e.dist;p[e.to] = G[u][i];if (!inq[e.to]) {Q.push(e.to);inq[e.to] = 1;if (++cnt[e.to] > n)return 1;}}}}return 0;}};BellmanFord solver;bool test(double x) {for (int i = 0; i < solver.m; i++)solver.edges[i].dist -= x;bool ret = solver.negativeCycle();for (int i = 0; i < solver.m; i++)solver.edges[i].dist += x;return ret;}int main() {int t;scanf("%d", &t);for (int cas = 1; cas <= t; cas++) {int n, m;scanf("%d%d", &n, &m);solver.init(n);int ub = 0;while (m--) {int u, v, w;scanf("%d%d%d", &u, &v, &w);u--, v--;ub = max(ub, w);solver.AddEdge(u, v, w);}printf("Case #%d: ", cas);if (!test(ub+1))printf("No cycle found.\n");else {double L = 0, R = ub;while (R-L > 1e-3) {double M = L + (R-L)/2;if (test(M))R = M;else L = M;}printf("%.2lf\n", L);}}return 0;}