I I U P C 2 0 0 6 | |
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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 | |
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Input | |
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The first line of input gives the number of | |
Output | |
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For each test case output one line containing “Case | |
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Constraints | |
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- n - a, - c | |
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Sample Input |
Output for Sample Input |
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2 |
Case #1: No cycle found. |
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Problemsetter: Mohammad Tavakoli Ghinani Alternate Solution: Cho | |
二分答案，判断是否有负权回路。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <queue>
6
7 using namespace std;
8
9 const int MAX_N = 100;
10 const double eps = 1e-5;
11 const int edge = 3000;
12 int first[MAX_N],Next[edge],v[edge];
13 double w[edge];
14 bool inq[MAX_N];
15 int cnt[MAX_N];
16 double d[MAX_N];
17 int N,M;
18 double sum = 0;
19
20 void add_edge(int id,int u) {
21 int e = first[u];
22 Next[id] = e;
23 first[u] = id;
24 }
25
26 bool bellman(double x) {
27 queue<int> q;
28 memset(inq,0,sizeof(inq));
29 memset(cnt,0,sizeof(cnt));
30 for(int i = 1; i <= N; ++i) {
31 d[i] = 0;
32 inq[i] = 1;
33 q.push(i);
34 }
35
36 while(!q.empty()) {
37 int u = q.front(); q.pop();
38 inq[u] = 0;
39 for(int e = first[u]; e != -1; e = Next[e]) {
40 if(d[ v[e] ] > d[u] + w[e] - x) {
41 d[ v[e] ] = d[u] + w[e] - x;
42 if(!inq[ v[e] ]) {
43 q.push( v[e] );
44 inq[ v[e] ] = 1;
45 if(++cnt[ v[e] ] > N) return true;
46 }
47 }
48 }
49 }
50
51 return false;
52
53 }
54
55 void solve() {
56 double l = 1,r = sum;
57 while(r - l >= eps) {
58 //printf("l = %f r = %f\n",l,r);
59 double mid = (l + r) / 2;
60 if(bellman(mid)) r = mid;
61 else l = mid;
62 }
63 if(bellman(sum + 1)) {
64 printf("%.2f\n",l);
65 } else {
66 printf("No cycle found.\n");
67 }
68 }
69
70 int main()
71 {
72 //freopen("sw.in","r",stdin);
73 int t;
74 scanf("%d",&t);
75 for(int ca = 1; ca <= t; ++ca) {
76 scanf("%d%d",&N,&M);
77 for(int i = 1; i <= N; ++i) first[i] = -1;
78 sum = 0;
79 for(int i = 0; i < M; ++i) {
80 int u;
81 scanf("%d%d%lf",&u,&v[i],&w[i]);
82 sum += w[i];
83 add_edge(i,u);
84 }
85
86 //printf("sum = %f\n",sum);
87 printf("Case #%d: ",ca);
88 solve();
89 }
90 //cout << "Hello world!" << endl;
91 return 0;
92 }