Graph Theory trainning-part-1 D. Going in cycle !!

 

D. Going in cycle !! Time Limit: 3000 msmemory limit: 131072kb64-bit integer Io format: % LLD Java class name: Main you are given a weighted directed graph NVertices and MEdges. 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. InputThe first line of input gives the number of cases, N. NTest Cases Follow. Each one starts with two numbers NAnd M. MLines follow, each has three positive number A, B, cWhich means there is an edge from vertex ATo BWith Weight C. OutputFor 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..

-N ≤ 50

-A, B ≤ n

-C ≤ 10000000

 

Sample Input

 

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

 

Output for sample input

Case #1: No cycle found.

Case #2: 2.50

 

Solution: Binary + spfa negative weight loop Determination

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <vector> 6 #include <climits> 7 #include <algorithm> 8 #include <cmath> 9 #include <queue>10 #define LL long long11 #define INF 0x3f3f3ff12 using namespace std;13 const int maxn = 60;14 const double exps = 1e-3;15 struct arc {16     int to;17     double w;18 };19 20 vector<arc>g[maxn];21 int cnt[maxn],n,m;22 double d[maxn];23 bool vis[maxn];24 bool spfa() {25     int i,j,v,u;26     queue<int>q;27     for(i = 1; i <= n; i++) {28         q.push(i);29         d[i] = INF;30         vis[i] = true;31         cnt[i] = 1;32     }33     d[1] = 0;34     while(!q.empty()) {35         u = q.front();36         q.pop();37         vis[u] = false;38         for(i = 0; i < g[u].size(); i++) {39             v = g[u][i].to;40             if(d[v] > d[u]+g[u][i].w) {41                 d[v] = d[u]+g[u][i].w;42                 if(!vis[v]) {43                     vis[v] = true;44                     cnt[v]++;45                     q.push(v);46                     if(cnt[u] > n) return true;47 48                 }49             }50         }51     }52     return false;53 }54 void modify(double val) {55     for(int i = 1; i <= n; i++) {56         for(int j = 0; j < g[i].size(); j++) {57             g[i][j].w += val;58         }59     }60 }61 bool test(double val) {62     modify(-val);63     bool it =  spfa();64     modify(val);65     return it;66 }67 int main() {68     int t,i,j,u,v,k = 1;69     double lt,rt,mid,w;70     scanf("%d",&t);71     while(t--) {72         scanf("%d%d",&n,&m);73         for(i = 0; i <= n; i++)74             g[i].clear();75         lt = INF;76         rt = 0;77         for(i = 0; i < m; i++) {78             scanf("%d%d%lf",&u,&v,&w);79             g[u].push_back((arc) {v,w});80             if(lt > w) lt = w;81             if(w > rt) rt = w;82         }83         printf("Case #%d: ",k++);84         if(test(rt+1.0)) {85             while(rt-lt > exps) {86                 mid = lt+(rt-lt)/2;87                 if(test(mid)) rt = mid;88                 else lt = mid;89             }90             printf("%.2f\n",rt);91         } else printf("No cycle found.\n");92     }93     return 0;94 }

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