图论trainning-part-1 D. Going in Cycle!!

D. Going in Cycle!!

Time Limit: 3000ms

Memory Limit: 131072KB

64-bit integer IO format: %lld      Java class name: Main

 

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 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.

 

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..

 

- 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

解题：二分+spfa负权回路判定

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <vector>
 #include <climits>
 #include <algorithm>
 #include <cmath>
 #include <queue>
 #define LL long long
 #define INF 0x3f3f3ff
 using namespace std;
 const int maxn = ;
 const double exps = 1e-;
 struct arc {
     int to;
     double w;
 };

 vector<arc>g[maxn];
 int cnt[maxn],n,m;
 double d[maxn];
 bool vis[maxn];
 bool spfa() {
     int i,j,v,u;
     queue<int>q;
     for(i = ; i <= n; i++) {
         q.push(i);
         d[i] = INF;
         vis[i] = true;
         cnt[i] = ;
     }
     d[] = ;
     while(!q.empty()) {
         u = q.front();
         q.pop();
         vis[u] = false;
         for(i = ; i < g[u].size(); i++) {
             v = g[u][i].to;
             if(d[v] > d[u]+g[u][i].w) {
                 d[v] = d[u]+g[u][i].w;
                 if(!vis[v]) {
                     vis[v] = true;
                     cnt[v]++;
                     q.push(v);
                     if(cnt[u] > n) return true;

                 }
             }
         }
     }
     return false;
 }
 void modify(double val) {
     for(int i = ; i <= n; i++) {
         for(int j = ; j < g[i].size(); j++) {
             g[i][j].w += val;
         }
     }
 }
 bool test(double val) {
     modify(-val);
     bool it =  spfa();
     modify(val);
     return it;
 }
 int main() {
     int t,i,j,u,v,k = ;
     double lt,rt,mid,w;
     scanf("%d",&t);
     while(t--) {
         scanf("%d%d",&n,&m);
         for(i = ; i <= n; i++)
             g[i].clear();
         lt = INF;
         rt = ;
         for(i = ; i < m; i++) {
             scanf("%d%d%lf",&u,&v,&w);
             g[u].push_back((arc) {v,w});
             if(lt > w) lt = w;
             if(w > rt) rt = w;
         }
         printf("Case #%d: ",k++);
         if(test(rt+1.0)) {
             while(rt-lt > exps) {
                 mid = lt+(rt-lt)/;
                 if(test(mid)) rt = mid;
                 else lt = mid;
             }
             printf("%.2f\n",rt);
         } else printf("No cycle found.\n");
     }
     return ;
 }