Description
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I I U P C 2 0 06 |
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Problem G: Going in Cycle!! |
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Input: standard input Output: standard output |
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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. |
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Input |
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The first line of input gives the number of cases, N. N test cases follow. Each one starts with two numbers n and m. m lines follow, each has three positive number a, b, c which means there is an edge from vertex a to b with weight of c. |
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Output |
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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.”. |
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Constraints |
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- n ≤ 50 - a, b ≤ n - c ≤ 10000000 |
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Sample Input |
Output for Sample Input |
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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;
}
UVA - 11090 Going in Cycle!! (Bellman-Ford算法判负环)
时间： 2024-10-17 12:11:30